CC Absolute Convergence and Error Bounds

Section 7.6 Absolute Convergence and Error Bounds

Motivating Questions

How well does the $n$ th partial sum of a convergent alternating series approximate the actual sum of the series? Why?
What is the difference between converging absolutely and converging conditionally?

When dealing with alternating series, there are a couple significant questions we would like to answer. First, can we approximate what the value of the sum of a convergent alternating series is? If so, how close is our estimate? Second, is there any connection between an alternating series and the related series with the similar series whose terms are all positive? In this section, we confront these questions.

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Subsection 7.6.1 Estimating Alternating Sums

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If the series converges, the argument for the Alternating Series Test also provides us with a method to determine how close the

n

th partial sum

S_{n}

is to the actual sum of the series. To see how this works, let

S

be the sum of a convergent alternating series, so

S = \sum_{k = 1}^{\infty} (- 1)^{k} a_{k} .

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Recall that the sequence of partial sums oscillates around the sum

S

so that

| S - S_{n} | < | S_{n + 1} - S_{n} | = a_{n + 1} .

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Therefore, the value of the term

a_{n + 1}

provides an error estimate for how well the partial sum

S_{n}

approximates the actual sum

S .

We summarize this fact in the statement of the Alternating Series Estimation Theorem.

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Alternating Series Estimation Theorem.

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Let

{a_{k}}

be a sequence of positive terms that decreases to

0

k \to \infty,

so that the alternating series

\sum_{k = 1}^{\infty} (- 1)^{k + 1} a_{k}

converges to a number

S .

S_{n} = \sum_{k = 1}^{n} (- 1)^{k + 1} a_{k}

is the

n

th partial sum of the alternating series, then

| S - S_{n} | \leq a_{n + 1} .

Example 7.41.

Determine how well the

100

th partial sum

S_{100}

\sum_{k = 1}^{\infty} \frac{(- 1)^{k + 1}}{k}

approximates the sum of the series.

Solution.

If we let

S

be the sum of the series

\sum_{k = 1}^{\infty} \frac{(- 1)^{k + 1}}{k},

then we know that

| S_{100} - S | < a_{101} .

Now

a_{101} = \frac{1}{101} \approx 0.0099,

so the 100th partial sum is within 0.0099 of the sum of the series. We have discussed the fact (and will later verify) that

S = \sum_{k = 1}^{\infty} \frac{(- 1)^{k + 1}}{k} = \ln (2),

and so

S \approx 0.693147

while

S_{100} = \sum_{k = 1}^{100} \frac{(- 1)^{k + 1}}{k} \approx 0.6881721793

We see that the actual difference between

S

and

S_{100}

is approximately

0.0049750013,

which is indeed less than

0.0099 .

Example 7.42.

Determine the number of terms it takes to approximate the sum of the convergent alternating series

\sum_{k = 1}^{\infty} \frac{(- 1)^{k + 1}}{k^{4}}

to within 0.0001.

Answer.

S_{10} \approx 0.9469925924;

\frac{7}{720} π^{4} \approx 0.9470328299 .

Solution.

First note that the sequence

{\frac{1}{k^{4}}}

decreases to 0, so the series

\sum_{k = 1}^{\infty} \frac{(- 1)^{k + 1}}{k^{4}}

converges by the Alternating Series Test. Let

S

be the sum

\sum_{k = 1}^{\infty} \frac{(- 1)^{k + 1}}{k^{4}}

and

S_{n}

the

n

th partial sum of the series. We know that

| S - S_{n} | < a_{n + 1} = \frac{1}{(n + 1)^{4}} .

So we only need to determine the value of

n

so that

\frac{1}{(n + 1)^{4}} < 0.0001 .

This happens when

(n + 1)^{4} < 10000

or when

n + 1 > 10 .

n = 10

will do. A computer algebra system gives

S_{10} \approx 0.9469925924

while

\frac{7}{720} π^{4} \approx 0.9470328299 .

These do agree to

0.0001 .

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Subsection 7.6.2 Absolute and Conditional Convergence

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A series such as

\begin{matrix} (7.16) & 1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \dots \end{matrix}

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whose terms are neither all nonnegative nor alternating is different from any series that we have considered so far. The behavior of such a series can be rather complicated, but there is an important connection between a series with some negative terms and series with all positive terms.

Example 7.43.

Explain why the series

$1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \dots$

must have a sum that is less than the series

$\sum_{k = 1}^{\infty} \frac{1}{k^{2}} .$
Explain why the series

$1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \dots$

must have a sum that is greater than the series

$\sum_{k = 1}^{\infty} - \frac{1}{k^{2}} .$
Given that the terms in the series

$1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \dots$

converge to 0, what do you think the previous two results tell us about the convergence status of this series?

Answer.

$1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \dots < \sum_{k = 1}^{\infty} \frac{1}{k^{2}} .$
$1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \dots > \sum_{k = 1}^{\infty} - \frac{1}{k^{2}} .$
We expect this series to converge to some finite number between $\sum_{k = 1}^{\infty} - \frac{1}{k^{2}}$ and $\sum_{k = 1}^{\infty} \frac{1}{k^{2}} .$

Solution.

Each term in the series

$1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \dots$

is of the form $\frac{1}{k^{2}}$ or $- \frac{1}{k^{2}} .$ Since each of these terms is less than or equal to $\frac{1}{k^{2}}$ it follows that

$1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \dots < \sum_{k = 1}^{\infty} \frac{1}{k^{2}}$

.
Each term in the series

$1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \dots$

is of the form $\frac{1}{k^{2}}$ or $- \frac{1}{k^{2}} .$ Since each of these terms is greater than or equal to $- \frac{1}{k^{2}}$ it follows that

$1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \dots > \sum_{k = 1}^{\infty} - \frac{1}{k^{2}}$

.
We know that $\sum_{k = 1}^{\infty} \frac{1}{k^{2}}$ is a $p$ -series with $p = 2 > 1$ and so is a convergent series. Since $\sum_{k = 1}^{\infty} \frac{1}{k^{2}} = - \sum_{k = 1}^{\infty} \frac{1}{k^{2}},$ the series $\sum_{k = 1}^{\infty} - \frac{1}{k^{2}}$ also converges. Since the terms in the series

$1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \dots$

converge to 0 and the series itself is bounded between two convergent series, we should expect this series to converge to some finite number between $\sum_{k = 1}^{\infty} - \frac{1}{k^{2}}$ and $\sum_{k = 1}^{\infty} \frac{1}{k^{2}} .$

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As the example in Example 7.43 suggests, if a series

\sum a_{k}

has some negative terms but

\sum | a_{k} |

converges, then the original series,

\sum a_{k},

must also converge. That is, if

\sum | a_{k} |

converges, then so must

\sum a_{k} .

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As we just observed, this is the case for the series (7.16), because the corresponding series of the absolute values of its terms is the convergent

p

-series

\sum \frac{1}{k^{2}} .

But there are series, such as the alternating harmonic series

\sum (- 1)^{k + 1} \frac{1}{k},

that converge while the corresponding series of absolute values,

\sum \frac{1}{k},

diverges. We distinguish between these behaviors by introducing the following language.

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Absolute vs. Conditional Convergence.

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Consider a series

\sum a_{k} .

The series $\sum a_{k}$ converges absolutely (or is absolutely convergent) provided that $\sum | a_{k} |$ converges.
The series $\sum a_{k}$ converges conditionally (or is conditionally convergent) provided that $\sum | a_{k} |$ diverges and $\sum a_{k}$ converges.

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In this terminology, the series (7.16) converges absolutely while the alternating harmonic series converges conditionally. Absolute convergence is a strong condition in that it implies convergence. That is, if the series

\sum | a_{k} |

converges, then the series

\sum a_{k}

converges as well. The converse is not true, as the alternating harmonic series shows.

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Absolute Convergence Implies Convergence.

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If a series

\sum_{k = 1}^{\infty} a_{k}

converges absolutely, then it converges in the usual sense. That is, if the series

\sum_{k = 1}^{\infty} | a_{k} |

converges, then

\sum_{k = 1}^{\infty} a_{k}

converges as well.

Example 7.44.

Consider the series $\sum (- 1)^{k} \frac{\ln (k)}{k} .$
1. Does this series converge? Explain.
2. Does this series converge absolutely? Explain what test you use to determine your answer.
Consider the series $\sum (- 1)^{k} \frac{\ln (k)}{k^{2}} .$
1. Does this series converge? Explain.
2. Does this series converge absolutely? Hint: Use the fact that $\ln (k) < \sqrt{k}$ for large values of $k$ and then compare to an appropriate $p$ -series.

Answer.

1. $\sum (- 1)^{k} \frac{\ln (k)}{k}$ converges.
2. $\sum (- 1)^{k} \frac{\ln (k)}{k}$ converges conditionally.
1. $\sum (- 1)^{k} \frac{\ln (k)}{k^{2}}$ converges.
2. $\sum (- 1)^{k} \frac{\ln (k)}{k^{2}}$ converges absolutely.

Solution.

1. By L’Hopital’s Rule we have
  
  $lim_{k \to \infty} \frac{\ln (k)}{k} = lim_{k \to \infty} \frac{1}{k} = 0 .$
  
  Also, $\frac{d}{d k} \frac{\ln (k)}{k} = \frac{1 - \ln (k)}{k^{2}}$ is negative when $k > e,$ so the sequence ${\frac{\ln (k)}{k}}$ ultimately decreases to 0. Since the first few terms in a series are irrelevant to its convergence or divergence, we conclude that the series $\sum (- 1)^{k} \frac{\ln (k)}{k}$ converges by the Alternating Series test.
2. Note that
  
  $\begin{aligned} lim_{t \to \infty} \int_{1}^{t} \frac{\ln (x)}{x} & = lim_{t \to \infty} {\frac{\ln (x)^{2}}{2} |}_{1}^{t} \\ = lim_{t \to \infty} \frac{\ln (t)^{2}}{2} \\ = lim_{t \to \infty} \frac{\ln (t)^{2}}{2} \\ = \infty . \end{aligned}$
  
  Since the improper integral diverges, the Integral Test shows that the series $\sum (- 1)^{k} \frac{\ln (k)}{k}$ diverges. So the series $\sum (- 1)^{k} \frac{\ln (k)}{k}$ converges conditionally.
1. By L’Hopital’s Rule we have
  
  $lim_{k \to \infty} \frac{\ln (k)}{k^{2}} = lim_{k \to \infty} \frac{1}{2 k^{2}} = 0 .$
  
  Also,
  
  $\frac{d}{d k} \frac{\ln (k)}{k^{2}} = \frac{k - 2 k \ln (k)}{k^{4}} = \frac{1 - 2 \ln (k)}{k^{3}}$
  
  is negative when $k > e,$ so the sequence ${\frac{\ln (k)}{k^{2}}}$ ultimately decreases to 0. Since the first few terms in a series are irrelevant to its convergence or divergence, we conclude that the series $\sum (- 1)^{k} \frac{\ln (k)}{k^{2}}$ converges by the Alternating Series test.
2. Notice that
  
  $lim_{k \to \infty} \frac{\ln (k)}{k^{1 / 2}} = lim_{k \to \infty} \frac{2}{k^{1 / 2}} = 0,$
  
  So $\frac{1}{\sqrt{k}}$ dominates $\ln (k)$ and $\ln (k) < \sqrt{k}$ for large $k .$ It follows that
  
  $\frac{\ln (k)}{k^{2}} < \frac{\sqrt{k}}{k^{2}} = \frac{1}{k^{3 / 2}}$
  
  for large $k .$ Therefore,
  
  $\sum \frac{\ln (k)}{k^{2}} < \sum \frac{1}{k^{3 / 2}}$
  
  for large $k .$ Since $\sum \frac{1}{k^{3 / 2}}$ is a $p$ -series with $p = \frac{3}{2} > 1,$ the series $\sum \frac{1}{k^{3 / 2}}$ converges. This forces the series $\sum \frac{\ln (k)}{k^{2}}$ to converge as well, by the direct comparison test. So $\sum (- 1)^{k} \frac{\ln (k)}{k^{2}}$ converges absolutely.

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Conditionally convergent series turn out to be very interesting. If the sequence

{a_{n}}

decreases to 0, but the series

\sum a_{k}

diverges, the conditionally convergent series

\sum (- 1)^{k} a_{k}

is right on the borderline of being a divergent series. As a result, any conditionally convergent series converges very slowly. Furthermore, some very strange things can happen with conditionally convergent series, as illustrated in some of the exercises.

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Subsection 7.6.3 Summary of Tests for Convergence of Series

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Convergence Test Summary.

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We have discussed several tests for convergence/divergence of series in our sections and in exercises. We close this section of the text with a summary of all the tests we have encountered, followed by an example that challenges you to decide which convergence test to apply to several different series.

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Geometric Series

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The geometric series

\sum a r^{k}

with ratio

r

converges for

- 1 < r < 1

and diverges for

| r | \geq 1 .

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The sum of the convergent geometric series

\sum_{k = 0}^{\infty} a r^{k}

\frac{a}{1 - r} .

Divergence Test

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If the sequence

a_{n}

does not converge to 0, then the series

\sum a_{k}

diverges.

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This is the first test to apply because the conclusion is simple. However, if

lim_{n \to \infty} a_{n} = 0,

no conclusion can be drawn.

Integral Test

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Let

f

be a positive, decreasing function on an interval

[c, \infty)

and let

a_{k} = f (k)

for each positive integer

k \geq c .

If $\int_{c}^{\infty} f (t) d t$ converges, then $\sum a_{k}$ converges.
If $\int_{c}^{\infty} f (t) d t$ diverges, then $\sum a_{k}$ diverges.

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Use this test when

f (x)

is easy to integrate.

Direct Comp. Test

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(see Ex 4 in Section 7.3)

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Let

0 \leq a_{k} \leq b_{k}

for each positive integer

k .

If $\sum b_{k}$ converges, then $\sum a_{k}$ converges.
If $\sum a_{k}$ diverges, then $\sum b_{k}$ diverges.

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Use this test when you have a series with known behavior that you can compare to — this test can be difficult to apply.

Limit Comp. Test

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Let

a_{n}

and

b_{n}

be sequences of positive terms. If

lim_{k \to \infty} \frac{a_{k}}{b_{k}} = L

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for some positive finite number

L,

then the two series

\sum a_{k}

and

\sum b_{k}

either both converge or both diverge.

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Easier to apply in general than the comparison test, but you must have a series with known behavior to compare. Useful to apply to series of rational functions.

Ratio Test

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Let

a_{k} \neq 0

for each

k

and suppose

lim_{k \to \infty} \frac{| a_{k + 1} |}{| a_{k} |} = r

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If $r < 1,$ then the series $\sum a_{k}$ converges absolutely.
If $r > 1,$ then the series $\sum a_{k}$ diverges.
If $r = 1,$ then test is inconclusive.

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This test is useful when a series involves factorials and powers.

Root Test

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(see Exercise 2 in Section 7.3)

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Let

a_{k} \geq 0

for each

k

and suppose

lim_{k \to \infty} \sqrt[k]{a_{k}} = r .

If $r < 1,$ then the series $\sum a_{k}$ converges.
If $r > 1,$ then the series $\sum a_{k}$ diverges.
If $r = 1,$ then test is inconclusive.

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In general, the Ratio Test can usually be used in place of the Root Test. However, the Root Test can be quick to use when

a_{k}

involves

k

th powers.

Alt. Series Test

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a_{n}

is a positive, decreasing sequence so that

lim_{n \to \infty} a_{n} = 0,

then the alternating series

\sum (- 1)^{k + 1} a_{k}

converges.

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This test applies only to alternating series — we assume that the terms

a_{n}

are all positive and that the sequence

{a_{n}}

is decreasing.

Alt. Series Est.

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Let

S_{n} = \sum_{k = 1}^{n} (- 1)^{k + 1} a_{k}

be the

n

th partial sum of the alternating series

\sum_{k = 1}^{\infty} (- 1)^{k + 1} a_{k} .

Assume

a_{n} > 0

for each positive integer

n,

the sequence

a_{n}

decreases to 0 and

lim_{n \to \infty} S_{n} = S .

Then it follows that

| S - S_{n} | < a_{n + 1} .

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This bound can be used to determine the accuracy of the partial sum

S_{n}

as an approximation of the sum of a convergent alternating series.

Example 7.45.

For (a)-(j), use appropriate tests to determine the convergence or divergence of the following series. Throughout, if a series is a convergent geometric series, find its sum.

$\sum_{k = 3}^{\infty} \frac{2}{\sqrt{k - 2}}$
$\sum_{k = 1}^{\infty} \frac{k}{1 + 2 k}$
$\sum_{k = 0}^{\infty} \frac{2 k^{2} + 1}{k^{3} + k + 1}$
$\sum_{k = 0}^{\infty} \frac{100^{k}}{k!}$
$\sum_{k = 1}^{\infty} \frac{2^{k}}{5^{k}}$
$\sum_{k = 1}^{\infty} \frac{k^{3} - 1}{k^{5} + 1}$
$\sum_{k = 2}^{\infty} \frac{3^{k - 1}}{7^{k}}$
$\sum_{k = 2}^{\infty} \frac{1}{k^{k}}$
$\sum_{k = 1}^{\infty} \frac{(- 1)^{k + 1}}{\sqrt{k + 1}}$
$\sum_{k = 2}^{\infty} \frac{1}{k \ln (k)}$
Determine a value of $n$ so that the $n$ th partial sum $S_{n}$ of the alternating series $\sum_{n = 2}^{\infty} \frac{(- 1)^{n}}{\ln (n)}$ approximates the sum to within 0.001.

Answer.

$\sum_{k = 3}^{\infty} \frac{2}{\sqrt{k - 2}}$ diverges.
$\sum_{k = 1}^{\infty} \frac{k}{1 + 2 k}$ diverges.
$\sum_{k = 0}^{\infty} \frac{2 k^{2} + 1}{k^{3} + k + 1}$ diverges.
$\sum_{k = 0}^{\infty} \frac{100^{k}}{k!}$ converges.
$\sum_{k = 1}^{\infty} \frac{2^{k}}{5^{k}}$ is a geometric series with ratio $\frac{2}{5}$ and sum $\frac{2}{3} .$
$\sum_{k = 1}^{\infty} \frac{k^{3} - 1}{k^{5} + 1}$ converges.
$\sum_{k = 2}^{\infty} \frac{3^{k - 1}}{7^{k}}$ is a convergent geometric series with $a = \frac{3}{49}$ and $r = \frac{3}{7}$ and sum $\frac{3}{28} .$
$\sum_{k = 2}^{\infty} \frac{1}{k^{k}}$ converges.
$\sum_{k = 1}^{\infty} \frac{(- 1)^{k + 1}}{\sqrt{k + 1}}$ converges.
$\sum_{k = 2}^{\infty} \frac{1}{k \ln (k)}$ diverges.
$\sum_{k = 2}^{\infty} \frac{(- 1)^{k}}{\ln (k)}$ converges very slowly.

Solution.

For large values of $k,$ $\frac{2}{\sqrt{k - 2}}$ looks like $\frac{2}{\sqrt{k}} .$ We will use the Limit Comparison Test to compare these two series:

$\begin{aligned} lim_{k \to \infty} \frac{\frac{2}{\sqrt{k - 2}}}{\frac{2}{\sqrt{k}}} & = lim_{k \to \infty} \frac{\sqrt{k}}{\sqrt{k - 2}} \\ = lim_{k \to \infty} \sqrt{\frac{1}{1 - \frac{2}{k}}} \\ = 1 . \end{aligned}$

Since this limit is a positive constant, we know that the two series $\sum_{k = 3}^{\infty} \frac{2}{\sqrt{k - 2}}$ and $\sum_{k = 3}^{\infty} \frac{2}{\sqrt{k}}$ either both converge or both diverge. Now $\sum_{k = 3}^{\infty} \frac{2}{\sqrt{k}} = 2 \sum_{k = 3}^{\infty} \frac{1}{k^{1 / 2}}$ is constant times a $p$ -series with $p = \frac{1}{2},$ so $\sum_{k = 3}^{\infty} \frac{2}{\sqrt{k}}$ diverges. Therefore, $\sum_{k = 3}^{\infty} \frac{2}{\sqrt{k - 2}}$ diverges as well.
Note that

$lim_{k \to \infty} \frac{k}{1 + 2 k} = lim_{k \to \infty} \frac{k}{2 k} = \frac{1}{2},$

so the Divergence Test shows that $\sum_{k = 1}^{\infty} \frac{k}{1 + 2 k}$ diverges.
For large values of $k,$ $\frac{2 k^{2} + 1}{k^{3} + k + 1}$ looks like $\frac{2}{k} .$ We will use the Limit Comparison Test to compare these two series:

$\begin{aligned} lim_{k \to \infty} \frac{\frac{2 k^{2} + 1}{k^{3} + k + 1}}{\frac{2}{k}} & = lim_{k \to \infty} \frac{(2 k^{2} + 1) k}{2 (k^{3} + k + 1)} \\ = lim_{k \to \infty} \frac{2 + \frac{1}{k^{2}}}{2 + \frac{1}{k^{2}} + \frac{1}{k^{3}}} \\ = 1 . \end{aligned}$

Since this limit is a positive constant, we know that the two series $\sum_{k = 1}^{\infty} \frac{2 k^{2} + 1}{k^{3} + k + 1}$ and $\sum_{k = 1}^{\infty} \frac{2}{k}$ either both converge or both diverge. Now $\sum_{k = 1}^{\infty} \frac{2}{k} = 2 \sum_{k = 1}^{\infty} \frac{1}{k}$ is constant times a $p$ -series with $p = 1,$ so $\sum_{k = 1}^{\infty} \frac{2}{k}$ diverges. Therefore, $\sum_{k = 0}^{\infty} \frac{2 k^{2} + 1}{k^{3} + k + 1}$ diverges as well.
Series that involve factorials are good candidates for the Ratio Test:

$\begin{aligned} lim_{k \to \infty} \frac{\frac{100^{k + 1}}{(k + 1)!}}{\frac{100^{k}}{k!}} & = lim_{k \to \infty} \frac{100^{k + 1} k!}{100^{k} (k + 1)!} \\ = lim_{k \to \infty} \frac{100}{k + 1} \\ = 0 . \end{aligned}$

Since this limit is 0, the Ratio Test tells us that the series $\sum_{k = 0}^{\infty} \frac{100^{k}}{k!}$ converges.
Notice that $\sum_{k = 1}^{\infty} \frac{2^{k}}{5^{k}} = \sum_{k = 1}^{\infty} {(\frac{2}{5})}^{k}$ is a geometric series with ratio $\frac{2}{5} .$ Let’s write out the first few terms to identify the value of $a :$

$\sum_{k = 1}^{\infty} {(\frac{2}{5})}^{k} = \frac{2}{5} + {(\frac{2}{5})}^{2} + {(\frac{2}{5})}^{3} + \dots = (\frac{2}{5}) [1 + \frac{2}{5} + {(\frac{2}{5})}^{2} + \dots] .$

So $a = \frac{2}{5} .$ Since the ratio of this geometric series is between $- 1$ and 1, the series converges. The sum of the series is

$\sum_{k = 1}^{\infty} \frac{2^{k}}{5^{k}} = (\frac{2}{5}) (\frac{1}{1 - \frac{2}{5}}) = (\frac{2}{5}) (\frac{5}{3}) = \frac{2}{3}$

.
For large values of $n,$ $\frac{k^{3} - 1}{k^{5} + 1}$ looks like $\frac{k^{3}}{k^{5}} = \frac{1}{k^{2}} .$ We will use the Limit Comparison Test to compare $\sum_{k = 1}^{\infty} \frac{k^{3} - 1}{k^{5} + 1}$ and $\sum_{k = 1}^{\infty} \frac{1}{k^{2}} :$

$\begin{aligned} lim_{k \to \infty} \frac{\frac{k^{3} - 1}{k^{5} + 1}}{\frac{1}{k^{2}}} & = lim_{k \to \infty} \frac{k^{5} - k^{2}}{k^{5} + 1} \\ = lim_{k \to \infty} \frac{1 - \frac{1}{k^{3}}}{1 + \frac{1}{k^{5}}} \\ = 1 . \end{aligned}$

Since this limit is 1, we know that the two series $\sum_{k = 1}^{\infty} \frac{k^{3} - 1}{k^{5} + 1}$ and $\sum_{k = 1}^{\infty} \frac{1}{k^{2}}$ either both converge or both diverge. Now $\sum_{k = 1}^{\infty} \frac{1}{k^{2}}$ is a $p$ -series with $p = 2,$ so $\sum_{k = 1}^{\infty} \frac{1}{k^{2}}$ converges. Therefore, $\sum_{k = 1}^{\infty} \frac{k^{3} - 1}{k^{5} + 1}$ converges as well.
This series looks geometric, but to be sure let’s write out the first few terms of this series:

$\sum_{k = 2}^{\infty} \frac{3^{k - 1}}{7^{k}} = \frac{3}{7^{2}} + \frac{3^{2}}{7^{3}} + \frac{3^{3}}{7^{4}} + \dots = (\frac{3}{7^{2}}) [1 + \frac{3}{7} + {(\frac{3}{7})}^{2} + {(\frac{3}{7})}^{3} + \dots] .$

So $\sum_{k = 2}^{\infty} \frac{3^{k - 1}}{7^{k}}$ is a geometric series with $a = \frac{3}{49}$ and $r = \frac{3}{7} .$ Since the ratio of this geometric series is between $- 1$ and 1, the series converges. The sum of the series is

$\begin{aligned} \sum_{k = 2}^{\infty} \frac{3^{k - 1}}{7^{k}} & = (\frac{3}{49}) (\frac{1}{1 - \frac{3}{7}}) \\ = (\frac{3}{49}) (\frac{7}{4}) \\ = \frac{3}{28} . \end{aligned}$
$\sum_{k = 2}^{\infty} \frac{1}{k^{k}}$ converges by direct comparison with the series $\sum_{k = 2}^{\infty} \frac{1}{k^{2}} .$ Note that for all $k \geq 2,$ we have $k^{k} \geq k^{2},$ so $\frac{1}{k^{k}} \leq \frac{1}{k^{2}} .$ Since $\sum_{k = 2}^{\infty} \frac{1}{k^{2}}$ is a convergent $p$ -series with $p = 2,$ then by the direct comparison test, $\sum_{k = 2}^{\infty} \frac{1}{k^{k}}$ also converges.
This series is an alternating series of the form $\sum_{k = 1}^{\infty} (- 1)^{k + 1} a_{k}$ with $a_{k} = \frac{1}{\sqrt{k + 1}} .$ Since

$lim_{k \to \infty} \frac{1}{\sqrt{k + 1}} = 0$

and $\frac{1}{\sqrt{k + 1}}$ decreases to 0, the Alternating Series Test shows that $\sum_{k = 1}^{\infty} \frac{(- 1)^{k + 1}}{\sqrt{k + 1}}$ converges.
For this example we will use the Integral Test with the substitution $w = \ln (x),$ $d w = \frac{1}{x} d x :$

$\begin{aligned} \int_{2}^{\infty} \frac{1}{x \ln (x)} d x & = lim_{b \to \infty} \int_{2}^{b} \frac{1}{x \ln (x)} d x \\ = lim_{b \to \infty} \int_{\ln (2)}^{\ln (b)} \frac{1}{w} d w \\ = lim_{b \to \infty} \ln (w) |_{\ln (2)}^{\ln (b)} \\ = lim_{b \to \infty} [\ln (\ln (b)) - \ln (\ln (2))] \\ = \infty . \end{aligned}$

Since $\int_{2}^{\infty} \frac{1}{x \ln (x)} d x$ diverges, we conclude $\sum_{k = 2}^{\infty} \frac{1}{k \ln (k)}$ diverges.
First note that the terms $\frac{1}{\ln (k)}$ decrease to $0,$ so the Alternating Series Test shows that the alternating series $\sum_{k = 2}^{\infty} \frac{(- 1)^{k}}{\ln (k)}$ converges. Let $S = \sum_{k = 2}^{\infty} \frac{(- 1)^{k}}{\ln (k)}$ and let $S_{n}$ be the $n$ th partial sum of this series. Recall that

$| S_{n} - S | < | S_{n} - S_{n + 1} | = a_{n + 1}$

where $a_{n + 1} = \frac{1}{\ln (n + 1)} .$ If we can find a value of $n$ so that $a_{n + 1} < 0.001,$ then $| S_{n} - S | < 0.001$ as desired. Now

$\begin{aligned} a_{n + 1} & < 0.001 \\ \frac{1}{\ln (n + 1)} & < 0.001 \\ \ln (n + 1) & > 1000 \\ n + 1 & > e^{1000} \\ n & > e^{1000} - 1 . \end{aligned}$

So if we choose $n = e^{1000},$ then $S_{n}$ approximates the sum of the series to within 0.001. Notice that $e^{1000} \approx 1.97 \times 10^{434}$ is a very large number. This shows that the series $\sum_{k = 2}^{\infty} \frac{(- 1)^{k}}{\ln (k)}$ converges very slowly.

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Subsection 7.6.4 Summary

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The difference between the $(n - 1)$ th partial sum $S_{n - 1}$ and the $n$ th partial sum $S_{n}$ of a convergent alternating series $\sum_{k = 1}^{\infty} (- 1)^{k} a_{k}$ is $| S_{n} - S_{n - 1} | = a_{n} .$ Since the partial sums oscillate around the sum $S$ of the series, it follows that

$| S - S_{n} | < a_{n} .$

So the $n$ th partial sum of a convergent alternating series $\sum_{k = 1}^{\infty} (- 1)^{k} a_{k}$ approximates the actual sum of the series to within $a_{n} .$
The convergence of an alternating series may relate to the convergence of the related series whose terms are all positive. Indeed, if an alternating series converges absolutely, then it also converges. However, an alternating series may converge even when the series with all positive terms diverges -- this is called conditional convergence.

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Exercises 7.6.5 Exercises

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1. Estimating the sum of an alternating series.

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For the following alternating series,

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\sum_{n = 1}^{\infty} a_{n} = 0.45 - \frac{(0.45)^{3}}{3!} + \frac{(0.45)^{5}}{5!} - \frac{(0.45)^{7}}{7!} + . . .

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how many terms do you have to compute in order for your approximation (your partial sum) to be within 0.0000001 from the convergent value of that series?

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2. Estimating the sum of a different alternating series.

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For the following alternating series,

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\sum_{n = 1}^{\infty} a_{n} = 1 - \frac{(0.4)^{2}}{2!} + \frac{(0.4)^{4}}{4!} - \frac{(0.4)^{6}}{6!} + \frac{(0.4)^{8}}{8!} - . . .

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how many terms do you have to go for your approximation (your partial sum) to be within 0.0000001 from the convergent value of that series?

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3. Estimating the sum of one more alternating series.

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For the following alternating series,

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\sum_{n = 1}^{\infty} a_{n} = 1 - \frac{1}{10} + \frac{1}{100} - \frac{1}{1000} + . . .

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how many terms do you have to go for your approximation (your partial sum) to be within 1e-06 from the convergent value of that series?

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4. A series that converges conditionally and slowly.

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Conditionally convergent series converge very slowly. As an example, consider the famous formula

⁸

We will derive this formula in upcoming work.

\begin{matrix} (7.17) & \frac{π}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots = \sum_{k = 0}^{\infty} (- 1)^{k} \frac{1}{2 k + 1} \end{matrix}

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In theory, the partial sums of this series could be used to approximate

π .

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Show that the series in (7.17) converges conditionally.
Let $S_{n}$ be the $n$ th partial sum of the series in (7.17). Calculate the error in approximating $\frac{π}{4}$ with $S_{100}$ and explain why this is not a very good approximation.
Determine the number of terms it would take in the series (7.17) to approximate $\frac{π}{4}$ to 10 decimal places. (The fact that it takes such a large number of terms to obtain even a modest degree of accuracy is why we say that conditionally convergent series converge very slowly.)

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5. A alternative approximation method for convergent alternating series.

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We have shown that if

\sum (- 1)^{k + 1} a_{k}

is a convergent alternating series, then the sum

S

of the series lies between any two consecutive partial sums

S_{n} .

This suggests that the average

\frac{S_{n} + S_{n + 1}}{2}

is a better approximation to

S

than is

S_{n} .

Show that $\frac{S_{n} + S_{n + 1}}{2} = S_{n} + \frac{1}{2} (- 1)^{n + 2} a_{n + 1} .$
Use this revised approximation in (a) with $n = 20$ to approximate $\ln (2)$ given that

$\ln (2) = \sum_{k = 1}^{\infty} (- 1)^{k + 1} \frac{1}{k} .$

Compare this to the approximation using just $S_{20} .$ For your convenience, $S_{20} = \frac{155685007}{232792560} .$

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6. A surprising result about the rearrangement of terms in conditionally convergent series.

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Conditionally convergent series exhibit interesting and unexpected behavior. In this exercise we examine the conditionally convergent alternating harmonic series

\sum_{k = 1}^{\infty} \frac{(- 1)^{k + 1}}{k}

and discover that addition is not commutative for conditionally convergent series. We will also encounter Riemann’s Theorem concerning rearrangements of conditionally convergent series. Before we begin, we remind ourselves that

\sum_{k = 1}^{\infty} \frac{(- 1)^{k + 1}}{k} = \ln (2),

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a fact which will be verified in a later section.

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First we make a quick analysis of the positive and negative terms of the alternating harmonic series.
1. Show that the series $\sum_{k = 1}^{\infty} \frac{1}{2 k}$ diverges.
2. Show that the series $\sum_{k = 1}^{\infty} \frac{1}{2 k + 1}$ diverges.
3. Based on the results of the previous parts of this exercise, what can we say about the sums $\sum_{k = C}^{\infty} \frac{1}{2 k}$ and $\sum_{k = C}^{\infty} \frac{1}{2 k + 1}$ for any positive integer $C ?$ Be specific in your explanation.
Recall addition of real numbers is commutative; that is

$a + b = b + a$

for any real numbers $a$ and $b .$ This property is valid for any sum of finitely many terms, but does this property extend when we add infinitely many terms together?

The answer is no, and something even more odd happens. Riemann’s Theorem (after the nineteenth-century mathematician Georg Friedrich Bernhard Riemann) states that a conditionally convergent series can be rearranged to converge to any prescribed sum. More specifically, this means that if we choose any real number $S,$ we can rearrange the terms of the alternating harmonic series $\sum_{k = 1}^{\infty} \frac{(- 1)^{k + 1}}{k}$ so that the sum is $S .$ To understand how Riemann’s Theorem works, let’s assume for the moment that the number $S$ we want our rearrangement to converge to is positive. Our job is to find a way to order the sum of terms of the alternating harmonic series to converge to $S .$
1. Explain how we know that, regardless of the value of $S,$ we can find a partial sum $P_{1}$
  
  $P_{1} = \sum_{k = 1}^{n_{1}} \frac{1}{2 k + 1} = 1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2 n_{1} + 1}$
  
  of the positive terms of the alternating harmonic series that equals or exceeds $S .$ Let
  
  $S_{1} = P_{1} .$
2. Explain how we know that, regardless of the value of $S_{1},$ we can find a partial sum $N_{1}$
  
  $N_{1} = - \sum_{k = 1}^{m_{1}} \frac{1}{2 k} = - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} - \dots - \frac{1}{2 m_{1}}$
  
  so that
  
  $S_{2} = S_{1} + N_{1} \leq S$
  
  .
3. Explain how we know that, regardless of the value of $S_{2},$ we can find a partial sum $P_{2}$
  
  $P_{2} = \sum_{k = n_{1} + 1}^{n_{2}} \frac{1}{2 k + 1} = \frac{1}{2 (n_{1} + 1) + 1} + \frac{1}{2 (n_{1} + 2) + 1} + \dots + \frac{1}{2 n_{2} + 1}$
  
  of the remaining positive terms of the alternating harmonic series so that
  
  $S_{3} = S_{2} + P_{2} \geq S$
  
  .
4. Explain how we know that, regardless of the value of $S_{3},$ we can find a partial sum
  
  $N_{2} = - \sum_{k = m_{1} + 1}^{m_{2}} \frac{1}{2 k} = - \frac{1}{2 (m_{1} + 1)} - \frac{1}{2 (m_{1} + 2)} - \dots - \frac{1}{2 m_{2}}$
  
  of the remaining negative terms of the alternating harmonic series so that
  
  $S_{4} = S_{3} + N_{2} \leq S$
  
  .
5. Explain why we can continue this process indefinitely and find a sequence ${S_{n}}$ whose terms are partial sums of a rearrangement of the terms in the alternating harmonic series so that $lim_{n \to \infty} S_{n} = S .$

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7. Determine whether a series is absolutely convergent, conditionally convergent, or divergent.

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Determine whether the following alternating series are absolutely convergent, conditionally convergent, or divergent. Answer "Absolutely Convergent", "Conditionally Convergent", or "Divergent".

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