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8.5: Alternating Series and Absolute Convergence - Mathematics


All of the series convergence tests we have used require that the underlying sequence ({a_n}) be a positive sequence. (We can relax this with Theorem 64 and state that there must be an (N>0) such that (a_n>0) for all (n>N); that is, ({a_n}) is positive for all but a finite number of values of (n).)

In this section we explore series whose summation includes negative terms. We start with a very specific form of series, where the terms of the summation alternate between being positive and negative.

Definition 34: alternating series

Let ({a_n}) be a positive sequence. An alternating series is a series of either the form

[sumlimits_{n=1}^infty (-1)^na_nqquad ext{or}qquad sumlimits_{n=1}^infty (-1)^{n+1}a_n.]

Recall the terms of Harmonic Series come from the Harmonic Sequence ({a_n} = {1/n}). An important alternating series is the Alternating Harmonic Series:

[sumlimits_{n=1}^infty (-1)^{n+1}dfrac1n = 1-dfrac12+dfrac13-dfrac14+dfrac15-dfrac16+cdots]

Geometric Series can also be alternating series when (r<0). For instance, if (r=-1/2), the geometric series is

[sumlimits_{n=0}^infty left(dfrac{-1}{2} ight)^n = 1-dfrac12+dfrac14-dfrac18+dfrac1{16}-dfrac1{32}+cdots]

Theorem 60 states that geometric series converge when (|r|<1) and gives the sum: ( sumlimits_{n=0}^infty r^n = dfrac1{1-r}). When (r=-1/2) as above, we find

[sumlimits_{n=0}^infty left(dfrac{-1}{2} ight)^n = dfrac1{1-(-1/2)} = dfrac 1{3/2} = dfrac23.]

A powerful convergence theorem exists for other alternating series that meet a few conditions.

theorem 70: alternating series test

Let ({a_n}) be a positive, decreasing sequence where ( limlimits_{n oinfty}a_n=0). Then

[sumlimits_{n=1}^infty (-1)^{n}a_n qquad ext{and}qquad sumlimits_{n=1}^infty (-1)^{n+1}a_n] converge.

The basic idea behind Theorem 70 is illustrated in Figure (PageIndex{1}). A positive, decreasing sequence ({a_n}) is shown along with the partial sums

[S_n = sumlimits_{i=1}^n(-1)^{i+1}a_i =a_1-a_2+a_3-a_4+cdots+(-1)^{n+1}a_n.]

Because ({a_n}) is decreasing, the amount by which (S_n) bounces up/down decreases. Moreover, the odd terms of (S_n) form a decreasing, bounded sequence, while the even terms of (S_n) form an increasing, bounded sequence. Since bounded, monotonic sequences converge (see Theorem 59) and the terms of ({a_n}) approach 0, one can show the odd and even terms of (S_n) converge to the same common limit (L), the sum of the series.

Example (PageIndex{1}): Applying the Alternating Series Test

Determine if the Alternating Series Test applies to each of the following series.

  1. ( sumlimits_{n=1}^infty (-1)^{n+1}dfrac{1}{n})
  2. ( sumlimits_{n=1}^infty (-1)^ndfrac{ln n}{n})
  3. (sumlimits_{n=1}^infty (-1)^{n+1}dfrac{|sin n|}{n^2})

Solution

  1. This is the Alternating Harmonic Series as seen previously. The underlying sequence is ({a_n} = {1/n}), which is positive, decreasing, and approaches 0 as (n oinfty). Therefore we can apply the Alternating Series Test and conclude this series converges.
    While the test does not state what the series converges to, we will see later that ( sumlimits_{n=1}^infty (-1)^{n+1}dfrac1n=ln2.)
  2. The underlying sequence is ({a_n} = {ln n/n}). This is positive and approaches 0 as (n oinfty) (use L'Hopital's Rule). However, the sequence is not decreasing for all (n). It is straightforward to compute (a_1=0), (a_2approx0.347), (a_3approx 0.366), and (a_4approx 0.347): the sequence is increasing for at least the first 3 terms.

    We do not immediately conclude that we cannot apply the Alternating Series Test. Rather, consider the long--term behavior of ({a_n}). Treating (a_n=a(n)) as a continuous function of (n) defined on ([1,infty)), we can take its derivative:[a^prime (n) = dfrac{1-ln n}{n^2}.] The derivative is negative for all (ngeq 3) (actually, for all (n>e)), meaning (a(n)=a_n) is decreasing on ([3,infty)). We can apply the Alternating Series Test to the series when we start with (n=3) and conclude that ( sumlimits_{n=3}^infty(-1)^ndfrac{ln n}{n}) converges; adding the terms with (n=1) and (n=2) do not change the convergence (i.e., we apply Theorem 64).

    The important lesson here is that as before, if a series fails to meet the criteria of the Alternating Series Test on only a finite number of terms, we can still apply the test.

  3. The underlying sequence is ({a_n} = |sin n|/n). This sequence is positive and approaches (0) as (n oinfty). However, it is not a decreasing sequence; the value of (|sin n|) oscillates between (0) and (1) as (n oinfty). We cannot remove a finite number of terms to make ({a_n}) decreasing, therefore we cannot apply the Alternating Series Test.

    Keep in mind that this does not mean we conclude the series diverges; in fact, it does converge. We are just unable to conclude this based on Theorem 70.

Key Idea 31 gives the sum of some important series. Two of these are

[sumlimits_{n=1}^infty dfrac1{n^2} =dfrac{pi^2}6 approx 1.64493 quad ext{and} quad sumlimits_{n=1}^infty dfrac{(-1)^{n+1}}{n^2} = dfrac{pi^2}{12}approx 0.82247.]

These two series converge to their sums at different rates. To be accurate to two places after the decimal, we need 202 terms of the first series though only 13 of the second. To get 3 places of accuracy, we need 1069 terms of the first series though only 33 of the second. Why is it that the second series converges so much faster than the first?

While there are many factors involved when studying rates of convergence, the alternating structure of an alternating series gives us a powerful tool when approximating the sum of a convergent series.

theorem 71: the alternating series approximation theorem

Let ({a_n}) be a sequence that satisfies the hypotheses of the Alternating Series Test, and let (S_n) and (L) be the (n^ ext{th}) partial sums and sum, respectively, of either ( sumlimits_{n=1}^infty (-1)^{n}a_n) or ( sumlimits_{n=1}^infty (-1)^{n+1}a_n). Then

  1. (|S_n-L| < a_{n+1}), and
  2. (L) is between (S_n) and (S_{n+1}).

Part 1 of Theorem 71 states that the (n^ ext{th}) partial sum of a convergent alternating series will be within (a_{n+1}) of its total sum. Consider the alternating series we looked at before the statement of the theorem, ( sumlimits_{n=1}^infty dfrac{(-1)^{n+1}}{n^2}). Since (a_{14} = 1/14^2 approx 0.0051), we know that (S_{13}) is within (0.0051) of the total sum.

Moreover, Part 2 of the theorem states that since (S_{13} approx 0.8252) and (S_{14}approx 0.8201), we know the sum (L) lies between (0.8201) and (0.8252). One use of this is the knowledge that (S_{14}) is accurate to two places after the decimal.

Some alternating series converge slowly. In Example (PageIndex{1}) we determined the series (sumlimits_{n=1}^infty (-1)^{n+1}dfrac{ln n}{n}) converged. With (n=1001), we find (ln n/n approx 0.0069), meaning that (S_{1000} approx 0.1633) is accurate to one, maybe two, places after the decimal. Since (S_{1001} approx 0.1564), we know the sum (L) is (0.1564leq Lleq0.1633).

Example (PageIndex{2}): Approximating the sum of convergent alternating series

Approximate the sum of the following series, accurate to within (0.001).

1.( sumlimits_{n=1}^infty (-1)^{n+1}dfrac{1}{n^3}qquad 2. sumlimits_{n=1}^infty (-1)^{n+1}dfrac{ln n}{n}).

Solution

  1. Using Theorem 71, we want to find (n) where (1/n^3 < 0.001): [egin{align*}dfrac1{n^3} &leq 0.001=dfrac{1}{1000} ^3 &geq 1000 &geq sqrt[3]{1000} &geq 10.end{align*}] Let (L) be the sum of this series. By Part 1 of the theorem, (|S_9-L|We can use Part 2 of the theorem to obtain an even more accurate result. As we know the (10^ ext{th}) term of the series is (-1/1000), we can easily compute (S_{10} = 0.901116). Part 2 of the theorem states that (L) is between (S_9) and (S_{10}), so (0.901116
  2. We want to find (n) where (ln (n)/n < 0.001). We start by solving (ln (n)/n = 0.001) for (n). This cannot be solved algebraically, so we will use Newton's Method to approximate a solution.

    Let (f(x) = ln(x)/x-0.001); we want to know where (f(x) = 0). We make a guess that (x) must be "large,'' so our initial guess will be (x_1=1000). Recall how Newton's Method works: given an approximate solution (x_n), our next approximation (x_{n+1}) is given by [x_{n+1} = x_n - dfrac{f(x_n)}{f^prime (x_n)}.]

    We find (f^prime (x) = ig(1-ln(x)ig)/x^2). This gives [egin{align*}x_2 &= 1000 - dfrac{ln(1000)/1000-0.001}{ig(1-ln(1000)ig)/1000^2} &= 2000.end{align*}]

    Using a computer, we find that Newton's Method seems to converge to a solution (x=9118.01) after 8 iterations. Taking the next integer higher, we have (n=9119), where (ln(9119)/9119 =0.000999903<0.001).

    Again using a computer, we find (S_{9118} = -0.160369). Part 1 of the theorem states that this is within (0.001) of the actual sum (L). Already knowing the 9,119(^ ext{th}) term, we can compute (S_{9119} = -0.159369), meaning (-0.159369 < L < -0.160369).

Notice how the first series converged quite quickly, where we needed only 10 terms to reach the desired accuracy, whereas the second series took over 9,000 terms.

One of the famous results of mathematics is that the Harmonic Series, ( sumlimits_{n=1}^infty dfrac1n) diverges, yet the Alternating Harmonic Series, ( sumlimits_{n=1}^infty (-1)^{n+1}dfrac1n), converges. The notion that alternating the signs of the terms in a series can make a series converge leads us to the following definitions.

Definition 35: absolute and conditional convergence

  1. A series ( sumlimits_{n=1}^infty a_n) converges absolutely if ( sumlimits_{n=1}^infty |a_n|) converges.
  2. A series ( sumlimits_{n=1}^infty a_n) converges conditionally if ( sumlimits_{n=1}^infty a_n) converges but ( sumlimits_{n=1}^infty |a_n|) diverges.

Thus we say the Alternating Harmonic Series converges conditionally.

Example (PageIndex{3}): Determining absolute and conditional convergence.

Determine if the following series converge absolutely, conditionally, or diverge.

1.( sumlimits_{n=1}^infty (-1)^ndfrac{n+3}{n^2+2n+5}qquad 2. sumlimits_{n=1}^infty (-1)^ndfrac{n^2+2n+5}{2^n}qquad 3.sumlimits_{n=3}^infty (-1)^ndfrac{3n-3}{5n-10})

Solution

  1. We can show the series [ sumlimits_{n=1}^infty left|(-1)^ndfrac{n+3}{n^2+2n+5} ight|= sumlimits_{n=1}^infty dfrac{n+3}{n^2+2n+5}]diverges using the Limit Comparison Test, comparing with (1/n).

    The series ( sumlimits_{n=1}^infty (-1)^ndfrac{n+3}{n^2+2n+5}) converges using the Alternating Series Test; we conclude it converges conditionally.

  2. We can show the series [ sumlimits_{n=1}^infty left|(-1)^ndfrac{n^2+2n+5}{2^n} ight|=sumlimits_{n=1}^infty dfrac{n^2+2n+5}{2^n}] converges using the Ratio Test.

    Therefore we conclude ( sumlimits_{n=1}^infty (-1)^ndfrac{n^2+2n+5}{2^n}) converges absolutely.

  3. The series [ sumlimits_{n=3}^infty left|(-1)^ndfrac{3n-3}{5n-10} ight| = sumlimits_{n=3}^infty dfrac{3n-3}{5n-10}]diverges using the (n^ ext{th}) Term Test, so it does not converge absolutely.

    The series ( sumlimits_{n=3}^infty (-1)^ndfrac{3n-3}{5n-10}) fails the conditions of the Alternating Series Test as ((3n-3)/(5n-10)) does not approach (0) as (n oinfty). We can state further that this series diverges; as (n oinfty), the series effectively adds and subtracts (3/5) over and over. This causes the sequence of partial sums to oscillate and not converge.

    Therefore the series ( sumlimits_{n=1}^infty (-1)^ndfrac{3n-3}{5n-10}) diverges.

Knowing that a series converges absolutely allows us to make two important statements, given in the following theorem. The first is that absolute convergence is "stronger'' than regular convergence. That is, just because ( sumlimits_{n=1}^infty a_n) converges, we cannot conclude that ( sumlimits_{n=1}^infty |a_n|) will converge, but knowing a series converges absolutely tells us that ( sumlimits_{n=1}^infty a_n) will converge.

One reason this is important is that our convergence tests all require that the underlying sequence of terms be positive. By taking the absolute value of the terms of a series where not all terms are positive, we are often able to apply an appropriate test and determine absolute convergence. This, in turn, determines that the series we are given also converges.

The second statement relates to rearrangements of series. When dealing with a finite set of numbers, the sum of the numbers does not depend on the order which they are added. (So (1+2+3 = 3+1+2).) One may be surprised to find out that when dealing with an infinite set of numbers, the same statement does not always hold true: some infinite lists of numbers may be rearranged in different orders to achieve different sums. The theorem states that the terms of an absolutely convergent series can be rearranged in any way without affecting the sum.

theorem 72: absolute convergence theorem

Let ( sumlimits_{n=1}^infty a_n) be a series that converges absolutely.

  1. ( sumlimits_{n=1}^infty a_n) converges.
  2. Let ({b_n}) be any rearrangement of the sequence ({a_n}). Then
    [sumlimits_{n=1}^infty b_n = sumlimits_{n=1}^infty a_n.]

In Example 8.5.3, we determined the series in part 2 converges absolutely. Theorem 72 tells us the series converges (which we could also determine using the Alternating Series Test).

The theorem states that rearranging the terms of an absolutely convergent series does not affect its sum. This implies that perhaps the sum of a conditionally convergent series can change based on the arrangement of terms. Indeed, it can. The Riemann Rearrangement Theorem (named after Bernhard Riemann) states that any conditionally convergent series can have its terms rearranged so that the sum is any desired value, including (infty)!

As an example, consider the Alternating Harmonic Series once more. We have stated that

[sumlimits_{n=1}^infty (-1)^{n+1}dfrac1n = 1-dfrac12+dfrac13-dfrac14+dfrac15-dfrac16+dfrac17cdots = ln 2,]

(see Key Idea 31 or Example 8.5.1).

Consider the rearrangement where every positive term is followed by two negative terms:

[1-dfrac12-dfrac14+dfrac13-dfrac16-dfrac18+dfrac15-dfrac1{10}-dfrac1{12}cdots]

(Convince yourself that these are exactly the same numbers as appear in the Alternating Harmonic Series, just in a different order.) Now group some terms and simplify:

[egin{align*}
left(1-dfrac12 ight)-dfrac14+left(dfrac13-dfrac16 ight)-dfrac18+left(dfrac15-dfrac1{10} ight)-dfrac1{12}+cdots &=
dfrac12-dfrac14+dfrac16-dfrac18+dfrac1{10}-dfrac{1}{12}+cdots &=
dfrac12left(1-dfrac12+dfrac13-dfrac14+dfrac15-dfrac16+cdots ight) & = dfrac12ln 2.
end{align*}]

By rearranging the terms of the series, we have arrived at a different sum! (One could try to argue that the Alternating Harmonic Series does not actually converge to (ln 2), because rearranging the terms of the series shouldn't change the sum. However, the Alternating Series Test proves this series converges to (L), for some number (L), and if the rearrangement does not change the sum, then (L = L/2), implying (L=0). But the Alternating Series Approximation Theorem quickly shows that (L>0). The only conclusion is that the rearrangement emph{did} change the sum.) This is an incredible result.

We end here our study of tests to determine convergence. The back cover of this text contains a table summarizing the tests that one may find useful.

While series are worthy of study in and of themselves, our ultimate goal within calculus is the study of Power Series, which we will consider in the next section. We will use power series to create functions where the output is the result of an infinite summation.


9.5 Alternating Series and Absolute Convergence

The series convergence tests we have used require that the underlying sequence < a n >be a positive sequence. (We can relax this with Theorem 9.2.5 and state that there must be an N > 0 such that a n > 0 for all n > N that is, < a n >is positive for all but a finite number of values of n .)

In this section we explore series whose summation includes negative terms. We start with a very specific form of series, where the terms of the summation alternate between being positive and negative.

Definition 9.5.1 Alternating Series

Let < b n >be a positive sequence. An alternating series is a series of either the form

∑ n = 1 ∞ ( - 1 ) n ⁢ b n or ∑ n = 1 ∞ ( - 1 ) n + 1 ⁢ b n .

We want to think that an alternating sequence < a n >is related to a positive sequence < b n >by a n = ( - 1 ) n ⁢ b n .

Recall the terms of Harmonic Series come from the Harmonic Sequence < b n >= < 1 / n >. An important alternating series is the Alternating Harmonic Series :

∑ n = 1 ∞ ( - 1 ) n + 1 ⁢ 1 n = 1 - 1 2 + 1 3 - 1 4 + 1 5 - 1 6 + ⋯

Geometric Series can also be alternating series when r < 0 . For instance, if r = - 1 / 2 , the geometric series is

∑ n = 0 ∞ ( - 1 2 ) n = 1 - 1 2 + 1 4 - 1 8 + 1 16 - 1 32 + ⋯

Theorem 9.2.1 states that geometric series converge when | r | < 1 and gives the sum: ∑ n = 0 ∞ r n = 1 1 - r . When r = - 1 / 2 as above, we find

∑ n = 0 ∞ ( - 1 2 ) n = 1 1 - ( - 1 / 2 ) = 1 3 / 2 = 2 3 .

A powerful convergence theorem exists for other alternating series that meet a few conditions.

Theorem 9.5.1 Alternating Series Test

Let < b n >be a positive, decreasing sequence where lim n → ∞ ⁡ b n = 0 . Then

∑ n = 1 ∞ ( - 1 ) n ⁢ b n and ∑ n = 1 ∞ ( - 1 ) n + 1 ⁢ b n

The basic idea behind Theorem 9.5.1 is illustrated in Figure 9.5.1 . A positive, decreasing sequence < b n >is shown along with the partial sums

S n = ∑ i = 1 n ( - 1 ) i + 1 ⁢ b i = b 1 - b 2 + b 3 - b 4 + ⋯ + ( - 1 ) n + 1 ⁢ b n .

Because < b n >is decreasing, the amount by which S n bounces up and down decreases. Moreover, the odd terms of S n form a decreasing, bounded sequence, while the even terms of S n form an increasing, bounded sequence. Since bounded, monotonic sequences converge (see Theorem 9.1.6 ) and the terms of < b n >approach 0, we will show below that the odd and even terms of S n converge to the same common limit L , the sum of the series.

S n Figure 9.5.1: Illustrating convergence with the Alternating Series Test.

Because < b n >is a decreasing sequence, we have b n - b n + 1 ≥ 0 . We will consider the even and odd partial sums separately. First consider the even partial sums.

S 2 = b 1 - b 2 ≥ 0 since ⁢ b 2 ≤ b 1
S 4 = b 1 - b 2 + b 3 - b 4 = S 2 + b 3 - b 4 ≥ S 2 since ⁢ b 3 - b 4 ≥ 0
S 6 = S 4 + b 5 - b 6 ≥ S 4 since ⁢ b 5 - b 6 ≥ 0
S 2 ⁢ n = S 2 ⁢ n - 2 + b 2 ⁢ n - 1 - b 2 ⁢ n ≥ S 2 ⁢ n - 2 since ⁢ b 2 ⁢ n - 1 - b 2 ⁢ n ≥ 0

0 ≤ S 2 ≤ S 4 ≤ S 6 ≤ ⋯ ≤ S 2 ⁢ n ≤ ⋯

so < S 2 ⁢ n >is an increasing sequence. But we can also write

S 2 ⁢ n = b 1 - b 2 + b 3 - b 4 + b 5 - ⋯ - b 2 ⁢ n - 2 + b 2 ⁢ n - 1 - b 2 ⁢ n
= b 1 - ( b 2 - b 3 ) - ( b 4 - b 5 ) - ⋯ - ( b 2 ⁢ n - 2 - b 2 ⁢ n - 1 ) - b 2 ⁢ n

Each term in parentheses is positive and b 2 ⁢ n is positive so we have S 2 ⁢ n ≤ b 1 for all n . We now have the sequence of even partial sums, < S 2 ⁢ n >, is increasing and bounded above so by Theorem 9.1.6 < S 2 ⁢ n >converges. Since we know it converges, we will assume it’s limit is L or

Next we determine the limit of the sequence of odd partial sums.

lim n → ∞ ⁡ S 2 ⁢ n + 1 = lim n → ∞ ⁡ ( S 2 ⁢ n + b 2 ⁢ n + 1 )
= lim n → ∞ ⁡ S 2 ⁢ n + lim n → ∞ ⁡ b 2 ⁢ n + 1
= L + 0
= L

Both the even and odd partial sums converge to L so we have lim n → ∞ ⁡ S n = L , which means the series is convergent. ∎

Watch the video:
Alternating Series — Another Example 4 from https://youtu.be/aOiZvfFAMW8

Example 9.5.1 Applying the Alternating Series Test

Determine if the Alternating Series Test applies to each of the following series.

1. ⁢ ∑ n = 1 ∞ ( - 1 ) n + 1 ⁢ 1 n 2. ⁢ ∑ n = 2 ∞ ( - 1 ) n ⁢ ln ⁡ n n 3. ⁢ ∑ n = 1 ∞ ( - 1 ) n + 1 ⁢ | sin ⁡ n | n 2

This is the Alternating Harmonic Series as seen previously. The underlying sequence is < b n >= < 1 / n >, which is positive, decreasing, and approaches 0 as n → ∞ . Therefore we can apply the Alternating Series Test and conclude this series converges.

While the test does not state what the series converges to, we will see later that ∑ n = 1 ∞ ( - 1 ) n + 1 ⁢ 1 n = ln ⁡ 2 .

The underlying sequence is < b n >= < ( ln ⁡ n ) / n >. This is positive for n ≥ 2 and lim n → ∞ ⁡ ln ⁡ n n = lim n → ∞ ⁡ 1 n = 0 (use L’Hôpital’s Rule). However, the sequence is not decreasing for all n . It is straightforward to compute b 1 ≈ 0.347 , b 2 ≈ 0.366 , and b 3 ≈ 0.347 : the sequence is increasing for at least the first 2 terms.

We do not immediately conclude that we cannot apply the Alternating Series Test. Rather, consider the long-term behavior of < b n >. Treating b n = b ⁢ ( n ) as a continuous function of n defined on [ 2 , ∞ ) , we can take its derivative:

The derivative is negative for all n ≥ 3 (actually, for all n > e ), meaning b ⁢ ( n ) = b n is decreasing on [ 3 , ∞ ) . We can apply the Alternating Series Test to the series when we start with n = 3 and conclude that ∑ n = 3 ∞ ( - 1 ) n ⁢ ln ⁡ n n converges adding the terms with n = 2 does not change the convergence (i.e., we apply Theorem 9.2.5 ).

The important lesson here is that as before, if a series fails to meet the criteria of the Alternating Series Test on only a finite number of terms, we can still apply the test.

The underlying sequence is < b n >= < | sin ⁡ n | / n 2 >. This sequence is positive and approaches 0 as n → ∞ . However, it is not a decreasing sequence the value of | sin ⁡ n | oscillates between 0 and 1 as n → ∞ . We cannot remove a finite number of terms to make < b n >decreasing, therefore we cannot apply the Alternating Series Test.

Keep in mind that this does not mean we conclude the series diverges in fact, it does converge. We are just unable to conclude this based on Theorem 9.5.1 .

One of the famous results of mathematics is that the Harmonic Series, ∑ n = 1 ∞ 1 n diverges, yet the Alternating Harmonic Series, ∑ n = 1 ∞ ( - 1 ) n + 1 ⁢ 1 n , converges. The notion that alternating the signs of the terms in a series can make a series converge leads us to the following definitions.

Definition 9.5.2 Absolute and Conditional Convergence

A series ∑ n = 1 ∞ a n converges absolutely if ∑ n = 1 ∞ | a n | converges.

A series ∑ n = 1 ∞ a n converges conditionally if ∑ n = 1 ∞ a n converges but ∑ n = 1 ∞ | a n | diverges.

Thus we say the Alternating Harmonic Series converges conditionally.

Example 9.5.2 Determining absolute and conditional convergence.

Determine if the following series converge absolutely, conditionally, or diverge.
1. ∑ n = 1 ∞ ( - 1 ) n ⁢ n + 3 n 2 + 2 ⁢ n + 5 2. ∑ n = 3 ∞ ( - 1 ) n ⁢ 3 ⁢ n - 3 5 ⁢ n - 10

∑ n = 1 ∞ | ( - 1 ) n ⁢ n + 3 n 2 + 2 ⁢ n + 5 | = ∑ n = 1 ∞ n + 3 n 2 + 2 ⁢ n + 5

diverges using the Limit Comparison Test, comparing with 1 / n .

The sequence < n + 3 n 2 + 2 ⁢ n + 5 >is monotonically decreasing, so that the series ∑ n = 1 ∞ ( - 1 ) n ⁢ n + 3 n 2 + 2 ⁢ n + 5 converges using the Alternating Series Test we conclude it converges conditionally.

∑ n = 3 ∞ | ( - 1 ) n ⁢ 3 ⁢ n - 3 5 ⁢ n - 10 | = ∑ n = 3 ∞ 3 ⁢ n - 3 5 ⁢ n - 10

diverges using the Test for Divergence, so it does not converge absolutely.

The series ∑ n = 3 ∞ ( - 1 ) n ⁢ 3 ⁢ n - 3 5 ⁢ n - 10 fails the conditions of the Alternating Series Test as ( 3 ⁢ n - 3 ) / ( 5 ⁢ n - 10 ) does not approach 0 as n → ∞ . We can state further that this series diverges as n → ∞ , the series effectively adds and subtracts 3 / 5 over and over. This causes the sequence of partial sums to oscillate and not converge.

Therefore the series ∑ n = 1 ∞ ( - 1 ) n ⁢ 3 ⁢ n - 3 5 ⁢ n - 10 diverges.

Knowing that a series converges absolutely allows us to make two important statements, given in the following theorem. The first is that absolute convergence is “stronger” than regular convergence. That is, just because ∑ n = 1 ∞ a n converges, we cannot conclude that ∑ n = 1 ∞ | a n | will converge, but knowing a series converges absolutely tells us that ∑ n = 1 ∞ a n will converge.

Theorem 9.5.2 Absolute Convergence Theorem

Let ∑ n = 1 ∞ a n be a series that converges absolutely.

Let < b n >be any rearrangement of the sequence < a n >. Then

One reason this is important is that our convergence tests all require that the underlying sequence of terms be positive. By taking the absolute value of the terms of a series where not all terms are positive, we are often able to apply an appropriate test and determine absolute convergence. This, in turn, determines that the series we are given also converges.

The second statement relates to rearrangements of series. When dealing with a finite set of numbers, the sum of the numbers does not depend on the order which they are added. (So 1 + 2 + 3 = 3 + 1 + 2 .) One may be surprised to find out that when dealing with an infinite set of numbers, the same statement does not always hold true: some infinite lists of numbers may be rearranged in different orders to achieve different sums. The theorem states that the terms of an absolutely convergent series can be rearranged in any way without affecting the sum.

The theorem states that rearranging the terms of an absolutely convergent series does not affect its sum. This implies that perhaps the sum of a conditionally convergent series can change based on the arrangement of terms. Indeed, it can. The Riemann Rearrangement Theorem (named after Bernhard Riemann) states that any conditionally convergent series can have its terms rearranged so that the sum is any desired value or infinity.

Before we consider an example, we state the following theorem that illustrates how the alternating structure of an alternating series is a powerful tool when approximating the sum of a convergent series.

Theorem 9.5.3 The Alternating Series Approximation Theorem

Let < b n >be a sequence that satisfies the hypotheses of the Alternating Series Test, and let S n and L be the n th partial sum and sum, respectively, of either ∑ n = 1 ∞ ( - 1 ) n ⁢ b n or ∑ n = 1 ∞ ( - 1 ) n + 1 ⁢ b n . Then

L is between S n and S n + 1 .

Part 1 of Theorem 9.5.3 states that the n th partial sum of a convergent alternating series will be within b n + 1 of its total sum. Consider the alternating series we looked at before the statement of the theorem, ∑ n = 1 ∞ ( - 1 ) n + 1 n 2 . Since b 14 = 1 / 14 2 ≈ 0.0051 , we know that S 13 is within 0.0051 of the total sum.

Moreover, Part 2 of the theorem states that since S 13 ≈ 0.8252 and S 14 ≈ 0.8201 , we know the sum L lies between 0.8201 and 0.8252 . One use of this is the knowledge that S 14 is accurate to two places after the decimal.

Some alternating series converge slowly. In Example 9.5.1 we determined the series ∑ n = 2 ∞ ( - 1 ) n + 1 ⁢ ln ⁡ n n converged. With n = 1001 , we find ( ln ⁡ n ) / n ≈ 0.0069 , meaning that S 1000 ≈ 0.1633 is accurate to one, maybe two, places after the decimal. Since S 1001 ≈ 0.1564 , we know the sum L is 0.1564 ≤ L ≤ 0.1633 .

Example 9.5.3 Approximating the sums of convergent alternating series

Approximate the sum of the following series, accurate to within 0.001 .

1. ⁢ ∑ n = 1 ∞ ( - 1 ) n + 1 ⁢ 1 n 3 2. ⁢ ∑ n = 1 ∞ ( - 1 ) n + 1 ⁢ ln ⁡ n n .

Using Theorem 9.5.3 , we want to find n where 1 / n 3 ≤ 0.001 :

1 n 3 ≤ 0.001 = 1 1000
n 3 ≥ 1000
n ≥ 1000 3
n ≥ 10 .

Let L be the sum of this series. By Part 1 of the theorem, | S 9 - L | < b 10 = 1 / 1000 . We can compute S 9 = 0.902116 , which our theorem states is within 0.001 of the total sum.

We can use Part 2 of the theorem to obtain an even more accurate result. As we know the 10 th term of the series is - 1 / 1000 , we can easily compute S 10 = 0.901116 . Part 2 of the theorem states that L is between S 9 and S 10 , so 0.901116 < L < 0.902116 .

We want to find n where ( ln ⁡ n ) / n ≤ 0.001 . We start by solving ( ln ⁡ n ) / n = 0.001 for n . This cannot be solved algebraically, so we will use Newton’s Method to approximate a solution.

Let f ⁢ ( x ) = ln ⁡ ( x ) / x - 0.001 we want to know where f ⁢ ( x ) = 0 . We make a guess that x must be “large,” so our initial guess will be x 1 = 1000 . Recall how Newton’s Method works: given an approximate solution x n , our next approximation x n + 1 is given by

x n + 1 = x n - f ⁢ ( x n ) f ′ ⁢ ( x n ) .

We find f ′ ⁢ ( x ) = ( 1 - ln ⁡ ( x ) ) / x 2 . This gives

x 2 = 1000 - ln ⁡ ( 1000 ) / 1000 - 0.001 ( 1 - ln ⁡ ( 1000 ) ) / 1000 2
= 2000 .

Using a computer, we find that Newton’s Method seems to converge to a solution x = 9118.01 after 8 iterations. Taking the next integer higher, we have n = 9119 , where ln ⁡ ( 9119 ) / 9119 = 0.000999903 < 0.001 .

Again using a computer, we find S 9118 = - 0.160369 . Part 1 of the theorem states that this is within 0.001 of the actual sum L . Already knowing the 9,119 th term, we can compute S 9119 = - 0.159369 , meaning

Notice how the first series converged quite quickly, where we needed only 10 terms to reach the desired accuracy, whereas the second series took over 9,000 terms.

We now consider the Alternating Harmonic Series once more. We have stated that

∑ n = 1 ∞ ( - 1 ) n + 1 ⁢ 1 n = 1 - 1 2 + 1 3 - 1 4 + 1 5 - 1 6 + 1 7 ⁢ ⋯ = ln ⁡ 2 ,

Consider the rearrangement where every positive term is followed by two negative terms:

1 - 1 2 - 1 4 + 1 3 - 1 6 - 1 8 + 1 5 - 1 10 - 1 12 ⁢ ⋯

(Convince yourself that these are exactly the same numbers as appear in the Alternating Harmonic Series, just in a different order.) Now group some terms and simplify:

( 1 - 1 2 ) - 1 4 + ( 1 3 - 1 6 ) - 1 8 + ( 1 5 - 1 10 ) - 1 12 + ⋯ =
1 2 - 1 4 + 1 6 - 1 8 + 1 10 - 1 12 + ⋯ =
1 2 ⁢ ( 1 - 1 2 + 1 3 - 1 4 + 1 5 - 1 6 + ⋯ ) = 1 2 ⁢ ln ⁡ 2 .

By rearranging the terms of the series, we have arrived at a different sum. (One could try to argue that the Alternating Harmonic Series does not actually converge to ln ⁡ 2 , because rearranging the terms of the series shouldn’t change the sum. However, the Alternating Series Test proves this series converges to L , for some number L , and if the rearrangement does not change the sum, then L = L / 2 , implying L = 0 . But the Alternating Series Approximation Theorem quickly shows that L > 0 . The only conclusion is that the rearrangement did change the sum.) This is an incredible result.

We mentioned earlier that the Integral Test did not work well with series containing factorial terms. The next section introduces the Ratio Test, which does handle such series well. We also introduce the Root Test, which is good for series where each term is raised to a power.


8.5: Alternating Series and Absolute Convergence - Mathematics

Be sure to check back, because this may change during the semester.

All numbers indicate sections from APEX Calculus, Version 3.0, and check the Errata for corrections to the text.

For Friday September 1 (Due 8/31 @ 8:00 pm)

Section 6.1 Substitution

  1. Substitution attempts to undo one of the techniques of differentiation. Which one is it?
  2. Use u-substitution to find an antiderivative of ( f(x) = 3x^2 cos(x^3) )
  3. Explain why ( dst int cos(x) sin(x)^2 dx) and ( dstint fracdx ) are essentially the same integral after performing a substitution.

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For Monday September 4

Labor Day. No class meeting or reading assignment due.

For Wednesday September 6 (Due 9/5 @ 8:00 pm)

Section 2.7 Derivatives of Inverse Functions

  1. Why do you think we are studying the inverse trig functions now?
  2. Find an antiderivative of ( f(x) = dst frac< 1 + x^6>)

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For Friday September 8 (Due 9/7 @ 8:00 pm)

Section 6.2 Integration by Parts

  1. Integration by parts attempts to undo one of the techniques of differentiation. Which one is it?
  2. Use integration by parts to find an antiderivative of (f(x) = 2x e^<4x>)

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For Monday September 11 (Due 9/10 @ 8:00 pm)

Section 6.2 Integration by Parts

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For Wednesday September 13 (Due 9/12 @ 8:00 pm)

Section 5.5 Numerical Integration

  1. Why would you ever want to numerically approximate an integral?
  2. Which would you expect to be MOST accurate: a Right Hand approximation, a Trapezoidal approximation, or a Simpson's approximation? Why?
  3. Which would you expect to be LEAST accurate: a Right Hand approximation, a Trapezoidal approximation, or a Simpson's approximation? Why?

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For Friday September 15 (Due 9/14 @ 8:00 pm)

Section 7.2 Volume by Cross-Sectional Area Disk and Washer

  1. Let R be the rectangle formed by the x-axis, the y-axis, and the lines y=1 and x=4. Describe the shape of the solid formed when R is rotated about the x-axis.
  2. Let T be the triangle formed by the lines y=2x, x=2 and the x-axis. Describe the shape of the solid formed when T is rotated about the line y = -1.

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For Monday September 18

Section 7.2 Volume by Cross-Sectional Area Disk and Washer

Re-read the section, but no Reading Questions for today

For Wednesday September 20 (Due 9/19 @ 8:00 pm)

Section 7.4 Arc Length and Surface Area

  1. Set up the integral that gives the length of the curve ( y=sin(2x)) from (x=0) to ( x=2pi).
  2. Set up the integral that gives the surface area of the surface formed when the curve ( y=x^2 + 2) from (x=0) to (x=3) is rotated about the x-axis.

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For Friday September 22 (Due 9/21 @ 8:00 pm)

Section 6.7 L'Hopital's Rule

  1. Does l'Hopital's Rule apply to ( dst lim_ frac) ? Why or why not?
  2. Does l'Hopital's Rule apply to ( dst lim_ frac)? Why or why not?
  3. For each limit in #1 and #2 where l'Hopital's applies, use it to find the limit.

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For Monday September 25 (Due 9/24 @ 8:00 pm)

Section 6.8 Improper Integration

  1. Explain why ( dstint_1^ frac<1>dx ) is improper.
  2. Explain why ( dstint_0^1 frac<1>dx ) is improper.
  3. Explain why ( dstint_<-1>^1 frac<1>dx ) is improper.

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For Wednesday September 27 (Due 9/26 @ 8:00 pm)

Section 6.8 Improper Integration

  1. If the improper integral ( int_1^ g(x) dx ) converges, what can you conclude about the improper integral ( int_1^ f(x) dx )?
  2. If the improper integral ( int_1^ f(x) dx ) diverges, what can you conclude about the improper integral ( int_1^ g(x) dx ) ?
  3. If the improper integral ( int_1^ f(x) dx ) converges, what can you conclude about the improper integral ( int_1^ g(x) dx ) ?

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For Friday September 29 (Due 9/28 @ 8:00 pm)

  1. Does the following sequence converge or diverge? Be sure to explain your answer.
    1, 3, 5, 7, 9, 11, 13, . . .
  2. Find a symbolic expression for the general term ak of the sequence 1, 2, 4, 8, 16, 32, . . .
  3. Is the following sequence bounded? Is it monotone? Explain. [ 1, -frac<1><2>, frac<1><4>, -frac<1><8>, frac<1><16>, -frac<1><32>, ldots ]

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For Monday October 2 (Due 10/1 @ 8:00 pm)

Section 8.2 Infinite Series

  1. There are two sequences associated with every series. What are they?
  2. Does the geometric series ( dst sum left( frac<1><4> ight)^k) converge or diverge? Why?
  3. Does the geometric series ( dst sum left( frac ight)^k) converge or diverge? Why?

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For Wednesday October 4 (Due 10/3 @ 8:00 pm)

Q & A for Exam 1. No Reading Assignment for today.

For Friday October 6 (Due 10/5 @ 8:00 pm)

Section 8.2 Infinite Series

  1. What does the n th -Term Theorem tell you about the series ( dst sum 2^k )?
  2. What does the n th -Term Theorem tell you about the series ( dst sum frac<1>)?

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For Monday October 9

Fall Break. No class meeting or reading assignment due.

For Wednesday October 11 (Due 10/10 @ 8:00 pm)

Section 8.3 Integral and Comparison Tests

  1. What does the Integral Test tell you about the series ( dst sum frac<1>)?
  2. What does the Integral Test tell you about the series ( dst sum frac<1>> )?
  3. What does the Direct Comparison Test tell you about the series ( dst sum frac<1>> )?

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For Friday October 13 (Due 10/12 @ 8:00 pm)

Section 8.3 Integral and Comparison Tests

  1. Use the Limit Comparison Test to show that ( dstsum frac<1>) converges.
  2. Explain why it would have been more difficult to apply the Direct Comparison Test to this series.

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For Monday October 16 (Due 10/15 @ 8:00 pm)

Section 8.5 Alternating Series and Absolute Convergence

  1. Why does this series converge?
  2. How closely does ( S_<50>), the 50th partial sum, approximate the value of the series? Why?

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For Wednesday October 18

Section 8.5 Alternating Series and Absolute Convergence

Re-read the section, but no Reading Questions for today

For Friday October 20 (Due 10/19 @ 8:00 pm)

Section 8.6 Power Series

  1. How do power series differ from the series we have looked at up to this point?
  2. What is the interval of convergence of a power series? Explain in your own words.

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For Monday October 23 (Due 10/22 @ 8:00 pm)

Section 8.7 Taylor Polynomials

    Explain the basic idea of constructing the n-th degree Taylor polynomial for a function f(x). Do not give the formula, but describe the idea in your own words in a few sentences.

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For Wednesday October 25 (Due 10/24 @ 8:00 pm)

Section 8.8 Taylor Series

  1. What is the difference between a Taylor series and a Maclaurin series?
  2. Why would you ever want to compute a Taylor series for a function like sin(x)?

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For Friday October 27 (Due 10/26 @ 8:00 pm)

Section 12.1 Introduction to Multivariable Functions

  1. Describe the level curves of the function (f(x,y)= x^2 - y)
  2. Describe the level surfaces of the function (g(x,y,z)=x^2+y^2+z^2)

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For Monday October 30 (Due 10/29 @ 8:00 pm)

Section 12.3 Partial Derivatives

  1. For a function (f(x,y)), what information does ( f_x(2,3)) give?
  2. How many second-order partial derivatives does a function (g(x,y)) have? Why?

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For Wednesday November 1 (Due 10/31 @ 8:00 pm)

Q & A for Exam 2. No Reading Assignment for today.

For Friday November 3

Moving Monday's class (10/30) here due to power outage on campus.

For Monday November 6 (Due 11/5 @ 8:00 pm)

Section 10.2 An Introduction to Vectors
Section 10.3 The Dot Product

  1. Give the unit vector in the same direction as ( vec<,v>=langle 2,3 angle)
  2. If the dot product ( vec<,u>cdot vec<,v>=0), what does this tell you about the vectors?

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For Wednesday November 8 (Due 11/7 @ 8:00 pm)

Section 12.6 Directional Derivatives

  1. What does the directional derivative ( D_> f(a,b)) measure?
  2. If (f(x,y) = 3xy^2 + 2x-4y^2), what is ( abla f(x,y)) ?

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For Thursday November 9

Section 12.8 Extreme Values

You should think about these questions while reading, but you do not need to submit answers.

  1. Where can the local extrema of a function f(x,y) occur?
  2. Why does the term "saddle point" make sense?

For Friday November 10

Section 12.8 Extreme Values

Re-read the section, but no Reading Questions for today

For Monday November 13 (Due 11/12 @ 8:00 pm)

Section 13.1 Iterated Integrals and Area

  1. Why would you want to calculate an iterated integral?
  2. Why would you want to switch the order of integration in an iterated integral?

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For Wednesday November 15 (Due 11/14 @ 8:00 pm)

Section 13.2 Double Integration and Volume

  1. If (f(x,y)) is a function of two variables and (R) is a rectangle in the xy-plane, what does ( intint_R f(x,y), dA) measure?
  2. Explain the idea of Fubini's Theorem in a couple of sentences in your own words.

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For Friday November 17

No class. I will be at a conference.

For Monday November 20

Section 13.2 Double Integration and Volume

Re-read the section, but no Reading Questions for today

For Wednesday November 22

Thanksgiving Break. No class meetings or reading assignments due.

For Friday November 24

Thanksgiving Break. No class meetings or reading assignments due.

For Monday November 27 (Due 11/26 @ 8:00 pm)

Section 9.4 Introduction to Polar Coordinates

  1. What do the coordinates ( (r, heta)) in polar coordinates measure?
  2. Why do you think we are studying polar coordinates now?
  3. Is the graph of the polar function ( r = 4cos( heta) ) is the graph of a function y=f(x)? Explain.

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For Wednesday November 29 (Due 11/28 @ 8:00 pm)

Q & A for Exam 3. No Reading Assignment for today.

For Friday December 1 (Due 11/30 @ 8:00 pm)

Section 13.3 Double Integration with Polar Coordinates

  1. Describe the shape of a polar "rectangle."
  2. Why would you ever want to use polar coordinates to evaluate a double integral?

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For Monday December 4 (Due 12/3 @ 8:00 pm)

Section 13.3 Double Integration with Polar Coordinates

Re-read the section, but no Reading Questions for today

For Wednesday December 6 (Due 12/5 @ 8:00 pm)

Section 13.4 Center of Mass

  1. Why do we need to use calculus to compute center of mass?
  2. When using a double integral to calculate ( M_x), the moment about the x-axis, what is the reason for multiplying the density function ( delta(x,y)) by (y) rather than by ( x)?

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For Friday December 8

The BIG Picture for the semester. No Reading Assignment for today.

For Monday November XY (Due XY @ 8:00 pm)

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For Wednesday November XY (Due XY @ 8:00 pm)

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For Friday November XY (Due XY @ 8:00 pm)

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Text Website


Calculus II MAT 1505 Syllabus

7.1 Integration by Parts
7.2 Trigonometric Integrals (Optional)
7.3 Trigonometric Substitution (Optional)
7.4 Int. of Rational Fns by Partial Frac. (Optional)
7.7 Approximate Integration
7.8 Improper Integrals

Chapter 8: Further Applications of Integration

8.1 Arc Length
8.4 Applications to Economics and Biology
8.5 Probability

Chapter 11: Infinite Sequences and Series

11.1 Sequences
11.2 Series (emphasize geometric series)
11.3 The Integral Test & Estimates of Sums
11.4 The Comparison Test
11.5 Alternating Series
11.6 Absolute Convergence, Ratio & Root Tests
11.8 Power Series**
11.9 Representations of Functions as Power Series**
11.10 Taylor and Maclaurin Series**
11.11 Application of Taylor Polynomials
** emphasize these sections

Chapter 10: Parametric Equations and Polar Coordinates

10.1 Curves Defined by Parametric Equations
10.2 Calculus with Parametric Equations
10.3 Polar Coordinates
10.4 Areas and Lengths in Polar Coordinates

Chapter 9: Differential Equations (if time allows)

9.1 Modeling with Differential Equation (Optional)
9.2 Direction Fields & Euler's Method (Optional)
9.3 Separable Equations (Optional)
9.4 Models for Population Growth (Optional)

This material is covered over a 14 week (56 class hours) semester.

Faculty have the option of utilizing WebAssign, a web-based supplement to our textbook. This portal provides graded and practice homework problems, online quizzes, video instruction and an eBook. WebAssign registration requires an access code, which is included with the text purchased from the Villanova Bookstore.

Note: If you choose to rent/purcase a book that comes without a WebAssign code, you will be required to purchase the registration code separately if your instructor uses WebAssign.

Upon the start of each semester, students will received a second code, called a class enrollment code, from their instructor. This code enrolls them into the specific course set up by their instructor for that semester.

Computer Algebra System (CAS)

Instructors will be using Maple or a comparable Computer Algebra System in the course. Students will work with the software interface to integrate numerically and symbolically, to solve differential equations, to explore the convergence or divergence of a series, as well as understand Taylor series.


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Mathematics-I

  • 4.0 4.5 -->
  • Publisher : uLektz Learning Solutions Private Limited
  • Course Code : MA6151
  • Author : uLektz
  • University : Anna University, Tamil Nadu

This is a digital eBook. No physical copy of this ebook will be delivered. This ebook can be accessed only through this web platform or uLektz readers supported mobile app..

Topics
UNIT I MATRICES

1.1 Eigen values and Eigen vectors of a real matrix

1.2 Characteristic equation

1.3 Properties of eigenvalues and eigenvectors

1.4 Statement and applications of Cayley - Hamilton Theorem

1.5 Diagonalization of matrices

1.6 Reduction of a quadratic form to canonical form by orthogonal transformation

1.7 Nature of quadratic forms

UNIT II SEQUENCES AND SERIES

2.3 Series of positive terms

2.7 Series of positive and negative terms

2.8 Absolute and conditional convergence

UNIT III APPLICATIONS OF DIFFERENTIAL CALCULUS

3.1 Curvature in Cartesian co-ordinates

3.2 Centre and radius of curvature

3.6 Evolute as envelope of normals

UNIT IV DIFFERENTIAL CALCULUS OF SEVERAL VARIABLES

4.4 Differentiation of implicit functions

4.5 Jacobian and properties

4.6 Taylor‟s series for functions of two variables

4.7 Maxima and minima of functions of two variables

4.8 Lagrange‟s method of undetermined multipliers.

UNIT V MULTIPLE INTEGRALS

5.1 Double integrals in Cartesian and polar coordinates

5.2 Change of order of integration

5.3 Area enclosed by plane curves

5.4 Change of variables in double integrals

5.5 Area of a curved surface

Publisher Detail:
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Math 231/249, Honors Calculus II

A second course in calculus, focusing on sequences and series, but also covering techniques of integration, parametric equations, polar coordinates, and complex numbers. While covering the same basic material as the standard sections, this honors class does so in more detail, including some additional topics. As such, it is for those students who, regardless of their major, are particularly interested in, and excited by, mathematics. In addition, a score of 5 on the AP Calculus AB exam, or a grade of "A" in Math 220 or 221 is required for enrollment. Students who enroll in this course must also register for Math 249 Q1H, the "honors supplement", with CRN 32044.

Grading:

  • Weekly Homework (15%). Homework will be assigned during each lecture and due at the beginning of class each Wednesday. No late homework will be accepted however, your lowest homework grade will be dropped so you are effectively allowed one infinitely late assignment. Collaboration on homework is permitted, nay encouraged. However, you must write up your solutions individually and understand them completely. You may use a computer or calculator on the HW for experimentation and to check your answers, but may not refer to it directly in the solution, e.g. "by Mathematica" is not an acceptable justification for deriving one equation from another. (Also, computers and calculators will not be allowed on the exams, so it's best not to get too dependent on them.)
  • Three in-class exams (20% each). These will be closed-book, calculator-free exams, though you will be allowed to bring one piece of paper with handwritten formulas. They will be on Fridays, in particular, September 19, October 17, and November 14.
  • A final exam (25%) Our final exam is scheduled for Friday, December 12 from 1:30-4:30 pm.

Textbook

  • Smith and Minton, Calculus: Early Transcendental Functions, 3rd edition, McGraw Hill, 2006 or 2007.

We will be covering Chapters 6-9, so either the "Single Variable" or "Full" version of this book is fine. As to the future value of the longer version for those planning on taking Math 241 (Calculus III), the honors sections of that course do not use this text, though some, but not all, of the standard sections do.


8.5: Alternating Series and Absolute Convergence - Mathematics

Absolute and Conditional Convergence In
your own words, describe the difference between absolute
and conditional convergence of an alternating series.

Answer

An Alternating series is absolutely converges if absolute value of series converges
An Alternating series is conditionally converges if absolute value of series diverges

Topics

Calculus of a Single Variable

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Watch More Solved Questions in Chapter 9

Video Transcript

So here is a difference of absolute today. So absolutes. Conventions. This means the absolute value of secrets. Serious. I'm very juice. That's conditional. Convergence mean they're serious convergence. Come on, I just But he's Ah, absolutes if I do of the area's dime, mergers diverges.


Full Texts PDF : Calculus I (v2) Calculus II (v2)

Section 2.7 Derivatives of Inverse Functions

Section 6.2 Integration by Parts

Section 6.3 Trigonometric Integrals

Section 6.4 Trigonometric Substitution

Section 6.5 Partial Fraction Decomposition

Section 6.7 L'Hopital's Rule

Section 6.8 Improper Integration

Section 8.1 Sequences

Section 8.2 Infinite Series

Section 8.3 Integral and Comparison Tests

Section 8.4 Ratio and Root Tests

Section 8.5 Alternating Series and Absolute Convergence


Calculus Early Transcendentals: Integral & Multi-Variable Calculus for Social Sciences

In this chapter we have a closer look at so-called , which arise in the study of analytic functions. A power series is basically an infinite degree polynomial that represents some function. Since we know a lot more about polynomial functions than arbitrary functions, this allows us to readily differentiate, integrate, and approximate some functions using power series.

Subsection 6.8.1 Power Series

Recall that the sum of a geometric series can be expressed using the simple formula:

if (|x|lt 1 ext<,>) and that the series diverges when (|x|ge 1 ext<.>) At the time, we thought of (x) as an unspecified constant, but we could just as well think of it as a variable, in which case the series

is a function, namely, the function (k/(1-x) ext<,>) as long as (|x|lt 1 ext<:>) Looking at this from the opposite perspective, this means that the function (k/(1-x)) can be represented as the sum of an infinite series. Why would this be useful? While (k/(1-x)) is a reasonably easy function to deal with, the more complicated representation (sum kx^n) does have some advantages: it appears to be an infinite version of one of the simplest function types — a polynomial. Later on we will investigate some of the ways we can take advantage of this ‘infinite polynomial’ representation, but first we should ask if other functions can even be represented this way.

The geometric series has a special feature that makes it unlike a typical polynomial—the coefficients of the powers of (x) are all the same, namely (k ext<.>) We will need to allow more general coefficients if we are to get anything other than the geometric series.

Definition 6.63 . Power Series Centred Around Zero.

A is a series of the form

where the (a_n) are real numbers.

As we did in the section on sequences, we can think of the (a_n) as being a function (a(n)) defined on the non-negative integers.

It is important to remember that the (a_n) do not depend on (x ext<.>)

Example 6.64 . Power Series Convergence.

Determine the values of (x) for which the power series (dssum_^infty ) converges.

We can investigate convergence using the Ratio Test:

Thus when (|x|lt 1) the series converges and when (|x|>1) it diverges, leaving only two values in doubt. When (x=1) the series is the harmonic series and diverges when (x=-1) it is the alternating harmonic series (actually the negative of the usual alternating harmonic series) and converges. Thus, we may think of (dssum_^infty ) as a function from the interval ([-1,1)) to the real numbers.

A bit of thought reveals that the Ratio Test applied to a power series will always have the same nice form. In general, we will compute

assuming that (ds lim |a_|/|a_n|) exists:

Then the series converges if (L|x|lt 1 ext<,>) that is, if (|x|lt 1/L ext<,>) and diverges if (|x|>1/L ext<.>)

Only the two values (x=pm1/L) require further investigation.

The value (1/L) is called the .

Thus the series will always define a function on the interval ((-1/L,1/L) ext<,>) that perhaps will extend to one or both endpoints as well.

This interval is referred to as the .

This interval is essentially the domain of the power series .

Then no matter what value (x) takes the limit is (0 ext<.>)

The series converges for all (x) and the function is defined for all real numbers.

Then no matter what value (x) takes the limit is infinite.

The series converges only when (x=0 ext<.>)

We can make these ideas a bit more general. Consider the series

This looks a lot like a power series, but with ((x+2)^n) instead of (x^n ext<.>) Let's try to determine the values of (x) for which it converges. This is just a geometric series, so it converges when

So the interval of convergence for this series is ((-5,1) ext<.>) The centre of this interval is at (-2 ext<,>) which is at distance 3 from the endpoints, so the radius of convergence is 3, and we say that the series is centred at (-2 ext<.>)

Interestingly, if we compute the sum of the series we get

Multiplying both sides by 1/3 we obtain

which we recognize as being equal to

so we have two series with the same sum but different intervals of convergence.

This leads to the following definition:

Definition 6.65 . Power Series Centred Around (a).

A centred at (a) has the form

where the (a) and (a_n) are real numbers.

Note: The power series centred at zero given in Definition 6.63 is a special case of the above definition when (a=0 ext<.>)

Convergence of a Power Series.

Given a power series (sum a_n(x-a)^n) and its radius of convergence (R ext<,>) the series behaves in one of three ways:

The series converges absolutely for (x) with (|x-a| lt R ext<,>) it diverges for (x) with (|x-a| > R ext<,>) and at (x=a-R) and (x=a+R) further investigation is needed.


8.5: Alternating Series and Absolute Convergence - Mathematics

Math 132 Syllabus

Some sections will be covered only in part. For indication of what is to be covered and what not, see the parenthetical qualifications below.

Chapter 4 - Antiderivatives
4.10 Antiderivatives

Chapter 5 - Integrals
5.1 Area and distances
5.2 The definite integral
5.3 The Fundamental Theorem of Calculus
5.4 Indefinite integrals and the Net Change Theorem
5.5 The Substitution Rule

Chapter 6 - Applications of Integration
6.1 Area between curves
6.2 Volumes

Chapter 7 - Techniques of Integration
7.1 Integration by parts
7.2 Trigonometric integrals
7.3 Trigonometric substitution
7.4 Integration of rational functions by partial fractions [just the case of a rational function whose denominator can be factored as (x - r)(x - s) with distinct r, s]
7.7 Approximate integration (omit error bounds formulas)
7.8 Improper integrals

Chapter 11 - Infinite Sequences and Series
11.1 Sequences
11.2 Series
11.3 The Integral Test and estimates of sums
11.4 The comparison tests
11.5 Alternating series
11.6 Absolute convergence the Ratio and Root Tests (omit subsection Rearrangements)
11.7 Strategy for testing series (for review)
11.8 Power series
11.9 Representation of functions as power series
11.10 Taylor and Maclaurin series (omit subsection Multiplication and Division of Power Series)
11.12 Applications of Taylor Polynomials (only the subsection Approximating functions by Polynomials)

Chapter 8 - Applications of Integration
8.1 Arc Length
8.5 Probability - omitted for Fall 2007

Chapter 10 - Parametric Equations and Polar Coordinates
10.1 Curves defined by parametric equations
10.2 Calculus with parametric curves (only subsections Tangents and Arc Length not Area and Surface Area)
10.3 Polar coordinates
10.4 Areas and length in polar coordinates


Watch the video: - Alternating Series (October 2021).