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## Section7.4Alternating Series and Absolute Convergence

###### Motivating Questions
• What is an alternating series?

• Under what conditions does an alternating series converge? Why?

• 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?

So far, we've mostly considered series with exclusively nonnegative terms. Next, we consider series that have some negative terms. For instance, the geometric series

\begin{equation*} 2 - \frac{4}{3} + \frac{8}{9} - \cdots + 2 \left(-\frac{2}{3} \right)^n + \cdots\text{,} \end{equation*}

has $a = 2$ and $r = -\frac{2}{3}\text{,}$ so that every other term alternates in sign. This series converges to

\begin{equation*} S = \frac{a}{1-r} = \frac{2}{1- \left(-\frac{2}{3}\right)} = \frac{6}{5}\text{.} \end{equation*}

In Example7.35 and our following discussion, we investigate the behavior of similar series where consecutive terms have opposite signs.

###### Example7.35

Example7.20 showed how we can approximate the number $e$ with linear, quadratic, and other polynomial approximations. We use a similar approach in this example to obtain linear and quadratic approximations to $\ln(2)\text{.}$ Along the way, we encounter a type of series that is different than most of the ones we have seen so far. Throughout this example, let $f(x) = \ln(1+x)\text{.}$

1. Find the tangent line to $f$ at $x=0$ and use this linearization to approximate $\ln(2)\text{.}$ That is, find $L(x)\text{,}$ the tangent line approximation to $f(x)\text{,}$ and use the fact that $L(1) \approx f(1)$ to estimate $\ln(2)\text{.}$

2. The linearization of $\ln(1+x)$ does not provide a very good approximation to $\ln(2)$ since $1$ is not that close to $0\text{.}$ To obtain a better approximation, we alter our approach; instead of using a straight line to approximate $\ln(2)\text{,}$ we use a quadratic function to account for the concavity of $\ln(1+x)$ for $x$ close to $0\text{.}$ With the linearization, both the function's value and slope agree with the linearization's value and slope at $x=0\text{.}$ We will now make a quadratic approximation $P_2(x)$ to $f(x) = \ln(1+x)$ centered at $x=0$ with the property that $P_2(0) = f(0)\text{,}$ $P'_2(0) = f'(0)\text{,}$ and $P''_2(0) = f''(0)\text{.}$

1. Let $P_2(x) = x - \frac{x^2}{2}\text{.}$ Show that $P_2(0) = f(0)\text{,}$ $P'_2(0) = f'(0)\text{,}$ and $P''_2(0) = f''(0)\text{.}$ Use $P_2(x)$ to approximate $\ln(2)$ by using the fact that $P_2(1) \approx f(1)\text{.}$

2. We can continue approximating $\ln(2)$ with polynomials of larger degree whose derivatives agree with those of $f$ at $0\text{.}$ This makes the polynomials fit the graph of $f$ better for more values of $x$ around $0\text{.}$ For example, let $P_3(x) = x - \frac{x^2}{2}+\frac{x^3}{3}\text{.}$ Show that $P_3(0) = f(0)\text{,}$ $P'_3(0) = f'(0)\text{,}$ $P''_3(0) = f''(0)\text{,}$ and $P'''_3(0) = f'''(0)\text{.}$ Taking a similar approach to preceding questions, use $P_3(x)$ to approximate $\ln(2)\text{.}$

3. If we used a degree $4$ or degree $5$ polynomial to approximate $\ln(1+x)\text{,}$ what approximations of $\ln(2)$ do you think would result? Use the preceding questions to conjecture a pattern that holds, and state the degree $4$ and degree $5$ approximation.

Solution
1. The linearization of $f$ at $x=a$ is

\begin{equation*} f(a) + f'(a)(x-a)\text{,} \end{equation*}

so the linearization $P_1(x)$ of $f(x) = \ln(1+x)$ at $x=0$ is

\begin{equation*} P_1(x) = 0 + \frac{1}{1+0}(x-0) = x\text{.} \end{equation*}

Now

\begin{equation*} f(x) \approx P_1(x) \end{equation*}

for $x$ close to $0$ and so

\begin{equation*} \ln(2) = \ln(1+1) \approx P_1(1) = 1\text{.} \end{equation*}
1. The derivatives of $P_2$ and $f$ are

\begin{align*} P_2(x) \amp = x - \frac{x^2}{2} \amp f(x) \amp = \ln(1+x)\\ P'_2(x) \amp = 1-x \amp f'(x) \amp = \frac{1}{1+x}\\ P''_2(x) \amp = -1 \amp f''(x) \amp = -\frac{1}{(1+x)^2}\text{,} \end{align*}

and so the derivatives of $P_2$ and $f$ evaluated at 0 are

\begin{align*} P_2(0) \amp = 0 \amp f(0) \amp = \ln(1) = 0\\ P'_2(0) \amp = 1 \amp f'(0) \amp = \frac{1}{1+0} = 1\\ P''_2(0) \amp = -1 \amp f''(0) \amp= -\frac{1}{(1+0)^2} = -1\text{.} \end{align*}

Then

\begin{equation*} \ln(2) = \ln(1+1) \approx P_2(1) = 1 - \frac{1}{2} = \frac{1}{2}\text{.} \end{equation*}
2. The derivatives of $P_3$ and $f$ are

\begin{align*} P_3(x) \amp = x-\frac{x^2}{2}+\frac{x^3}{3} \amp f(x) \amp = \ln(1+x)\\ P'_3(x) \amp = 1 - x + x^2 \amp f'(x) \amp = \frac{1}{1+x}\\ P''_3(x) \amp = -1+2x \amp f''(x) \amp = -\frac{1}{(1+x)^2}\\ P'''_3(x) \amp = 2 \amp f'''(x) \amp = \frac{2}{(1+x)^3}\text{,} \end{align*}

and so the derivatives of $P_3$ and $f$ evaluated at 0 are

\begin{align*} P_3(0) \amp = 0 \amp f(0) \amp = \ln(1+0) = 0\\ P'_3(0) \amp = 1 \amp f'(0) \amp = \frac{1}{1+0} = 1\\ P''_3(0) \amp = -1 \amp f''(0) \amp = -\frac{1}{(1+0)^2} = -1\\ P'''_3(0) \amp = 2 \amp f'''(0) \amp = \frac{2}{(1+0)^3} = 2\text{.} \end{align*}

Then

\begin{equation*} \ln(2) = \ln(1+1) \approx P_3(0) = 1 - \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \approx 0.83\text{.} \end{equation*}

### SubsectionThe Alternating Series Test

Example7.35 gives us several approximations to $\ln(2)\text{.}$ The linear approximation is $1\text{,}$ and the quadratic approximation is $1 - \frac{1}{2} = \frac{1}{2}\text{.}$ If we continue this process, cubic, quartic (degree $4$), quintic (degree $5$), and higher degree polynomials give us the approximations to $\ln(2)$ in Table7.36.

The pattern here shows that $\ln(2)$ can be approximated by the partial sums of the infinite series

\begin{equation} \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k}\label{eq-ln-2}\tag{7.15} \end{equation}

where the alternating signs are indicated by the factor $(-1)^{k+1}\text{.}$ We call such a series an alternating series.

Using computational technology, we find that the sum of the first 100 terms in this series is 0.6881721793. As a comparison, $\ln(2) \approx 0.6931471806\text{.}$ This shows that even though the series(7.15) converges to $\ln(2)\text{,}$ it must do so quite slowly, since the sum of the first 100 terms isn't particularly close to $\ln(2)\text{.}$ We will investigate the issue of how quickly an alternating series converges later in this section.

###### Alternating Series

An alternating series is a series of the form

\begin{equation*} \sum_{k=0}^{\infty} (-1)^k a_k\text{,} \end{equation*}

where $a_k \gt 0$ for each $k\text{.}$

We have some flexibility in how we write an alternating series; for example, the series

\begin{equation*} \sum_{k=1}^{\infty} (-1)^{k+1} a_k\text{,} \end{equation*}

whose index starts at $k = 1\text{,}$ is also alternating. As we will soon see, there are several very nice results that hold for alternating series, while alternating series can also demonstrate some unusual behaivior.

It is important to remember that most of the series tests we have seen in previous sections apply only to series with nonnegative terms. Alternating series require a different test.

###### Example7.37

Remember that, by definition, a series converges if and only if its corresponding sequence of partial sums converges.

1. Calculate the first few partial sums (to 10 decimal places) of the alternating series

\begin{equation*} \sum_{k=1}^{\infty} (-1)^{k+1}\frac{1}{k}\text{.} \end{equation*}

Label each partial sum with the notation $S_n = \sum_{k=1}^{n} (-1)^{k+1}\frac{1}{k}$ for an appropriate choice of $n\text{.}$

2. Plot the sequence of partial sums from part (a). What do you notice about this sequence?

1.  $\sum_{k=1}^{1} (-1)^{k+1}\frac{1}{k}$ $1$ $\sum_{k=1}^{6} (-1)^{k+1}\frac{1}{k}$ $0.6166666667$ $\sum_{k=1}^{2} (-1)^{k+1}\frac{1}{k}$ $0.5$ $\sum_{k=1}^{7} (-1)^{k+1}\frac{1}{k}$ $0.7595238095$ $\sum_{k=1}^{3} (-1)^{k+1}\frac{1}{k}$ $0.8333333333$ $\sum_{k=1}^{8} (-1)^{k+1}\frac{1}{k}$ $0.6345238095$ $\sum_{k=1}^{4} (-1)^{k+1}\frac{1}{k}$ $0.5833333333$ $\sum_{k=1}^{9} (-1)^{k+1}\frac{1}{k}$ $0.7456349206$ $\sum_{k=1}^{5} (-1)^{k+1}\frac{1}{k}$ $0.7833333333$ $\sum_{k=1}^{10} (-1)^{k+1}\frac{1}{k}$ $0.6456349206$
2. There appears to be a limit for the sequence of partial sums.

Solution
1. The completed table is shown below.

 $\sum_{k=1}^{1} (-1)^{k+1}\frac{1}{k}$ $1$ $\sum_{k=1}^{6} (-1)^{k+1}\frac{1}{k}$ $0.6166666667$ $\sum_{k=1}^{2} (-1)^{k+1}\frac{1}{k}$ $0.5$ $\sum_{k=1}^{7} (-1)^{k+1}\frac{1}{k}$ $0.7595238095$ $\sum_{k=1}^{3} (-1)^{k+1}\frac{1}{k}$ $0.8333333333$ $\sum_{k=1}^{8} (-1)^{k+1}\frac{1}{k}$ $0.6345238095$ $\sum_{k=1}^{4} (-1)^{k+1}\frac{1}{k}$ $0.5833333333$ $\sum_{k=1}^{9} (-1)^{k+1}\frac{1}{k}$ $0.7456349206$ $\sum_{k=1}^{5} (-1)^{k+1}\frac{1}{k}$ $0.7833333333$ $\sum_{k=1}^{10} (-1)^{k+1}\frac{1}{k}$ $0.6456349206$
2. The entries in the sequence of partial sums from part (a) is shown in the figure below.

Notice that the partial sums seem to oscillate back and forth around some fixed number, getting closer to that fixed number at each successive step. So there appears to be a limit for this sequence of partial sums.

Example7.37 illustrates the general behavior of any convergent alternating series. We see that the partial sums of the alternating harmonic series oscillate around a fixed number that turns out to be the sum of the series.

Recall that if $\lim_{k \to \infty} a_k \neq 0\text{,}$ then the series $\sum a_k$ diverges by the Divergence Test. From this point forward, we will thus only consider alternating series

\begin{equation*} \sum_{k=1}^{\infty} (-1)^{k+1} a_k \end{equation*}

in which the sequence $a_k$ consists of positive numbers that decrease to $0\text{.}$ The $n$th partial sum $S_n$ is

\begin{equation*} S_n = \sum_{k=1}^n (-1)^{k+1} a_k\text{.} \end{equation*}

Notice that

• $S_2 = a_1 - a_2\text{,}$ and since $a_1 \gt a_2$ we have $0 \lt S_2 \lt S_1 \text{.}$

• $S_3 = S_2+a_3$ and so $S_2 \lt S_3\text{.}$ But $a_3 \lt a_2\text{,}$ so $S_3 \lt S_1\text{.}$ Thus, $0 \lt S_2 \lt S_3 \lt S_1 \text{.}$

• $S_4 = S_3-a_4$ and so $S_4 \lt S_3\text{.}$ But $a_4 \lt a_3\text{,}$ so $S_2 \lt S_4\text{.}$ Thus, $0 \lt S_2 \lt S_4 \lt S_3 \lt S_1 \text{.}$

• $S_5 = S_4+a_5$ and so $S_4 \lt S_5\text{.}$ But $a_5 \lt a_4\text{,}$ so $S_5 \lt S_3\text{.}$ Thus, $0 \lt S_2 \lt S_4 \lt S_5 \lt S_3 \lt S_1 \text{.}$

This pattern continues as illustrated in Figure7.39 (with $n$ odd) so that each partial sum lies between the previous two partial sums. Note further that the absolute value of the difference between the $(n-1)$st partial sum $S_{n-1}$ and the $n$th partial sum $S_n$ is

\begin{equation*} \left\lvert S_n - S_{n-1} \right\rvert = a_n\text{.} \end{equation*}

Because the sequence $\{a_n\}$ converges to $0\text{,}$ the distance between successive partial sums becomes as close to zero as we'd like, and thus the sequence of partial sums converges (even though we don't know the exact value to which it converges).

The preceding discussion has demonstrated the truth of the Alternating Series Test.

###### The Alternating Series Test

Given an alternating series $\sum (-1)^k a_k \text{,}$ if the sequence $\{a_k\}$ of positive terms decreases to 0 as $k \to \infty\text{,}$ then the alternating series converges.

Note that if the limit of the sequence $\{a_k\}$ is not 0, then the alternating series diverges.

###### Example7.40

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

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

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

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

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

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

Solution
1. Since $(k+1)^2 + 2 \gt k^2+2$ it follows that $\frac{1}{(k+1)^2+2} \lt \frac{1}{k^2+2}$ and the sequence $\frac{1}{k^2+2}$ decreases to 0. The Alternating Series Test shows then that the series $\sum_{k=1}^{\infty} \frac{(-1)^k}{k^2+2}$ converges.

2. In this case we have that $\lim_{k \to \infty} \frac{2k}{k+5} = 2$ and so the sequence of $k$th terms does not even converge to 0. So the series $\sum_{k=1}^{\infty} \frac{2(-1)^{k+1}k}{k+5}$ diverges by the Divergence Test.

3. We know that the natural log is an increasing function, so $\frac{1}{\ln(k)}$ is a decreasing function. It is also true that $\lim_{k \to \infty} \frac{1}{\ln(k)} = 0$ and so the series $\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\ln(k)}$ converges by the Alternating Series Test.

### SubsectionEstimating Alternating Sums

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

\begin{equation*} S = \sum_{k=1}^{\infty} (-1)^k a_k\text{.} \end{equation*}

Recall that the sequence of partial sums oscillates around the sum $S$ so that

\begin{equation*} \left|S - S_n \right| \lt \left| S_{n+1} - S_n \right| = a_{n+1}\text{.} \end{equation*}

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 $S\text{.}$ We summarize this fact in the statement of the Alternating Series Estimation Theorem.

###### Alternating Series Estimation Theorem

If the alternating series $\sum_{k=1}^{\infty} (-1)^{k+1}a_k$ converges and has sum $S\text{,}$ and $S_n = \sum_{k=1}^{n} (-1)^{k+1}a_k$ is the $n$th partial sum of the alternating series, then

\begin{equation*} \left\lvert \sum_{k=1}^{\infty} (-1)^{k+1}a_k - S_n \right\rvert \leq a_{n+1}\text{.} \end{equation*}
###### Example7.41

Determine how well the $100$th partial sum $S_{100}$ of

\begin{equation*} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \end{equation*}

approximates the sum of the series.

Solution

If we let $S$ be the sum of the series $\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\text{,}$ then we know that

\begin{equation*} \left| S_{100} - S \right| \lt a_{101}\text{.} \end{equation*}

Now

\begin{equation*} a_{101} = \frac{1}{101} \approx 0.0099\text{,} \end{equation*}

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

\begin{equation*} S = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = \ln(2)\text{,} \end{equation*}

and so $S \approx 0.693147$ while

\begin{equation*} S_{100} = \sum_{k=1}^{100} \frac{(-1)^{k+1}}{k} \approx 0.6881721793\text{.} \end{equation*}

We see that the actual difference between $S$ and $S_{100}$ is approximately $0.0049750013\text{,}$ which is indeed less than $0.0099\text{.}$

###### Example7.42

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

\begin{equation*} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^4} \end{equation*}

to within 0.0001.

$S_{10} \approx 0.9469925924\text{;}$ $\frac{7}{720} \pi^4 \approx 0.9470328299\text{.}$

Solution

First note that the sequence $\left\{\frac{1}{k^4}\right\}$ 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

\begin{equation*} |S - S_n| \lt a_{n+1} = \frac{1}{(n+1)^4}\text{.} \end{equation*}

So we only need to determine the value of $n$ so that

\begin{equation*} \frac{1}{(n+1)^4} \lt 0.0001\text{.} \end{equation*}

This happens when $(n+1)^4 \lt 10000$ or when $n+1 \gt 10\text{.}$ So $n = 10$ will do. A computer algebra system gives $S_{10} \approx 0.9469925924$ while $\frac{7}{720} \pi^4 \approx 0.9470328299\text{.}$ These do agree to $0.0001\text{.}$

### SubsectionAbsolute and Conditional Convergence

A series such as

\begin{equation} 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} + \cdots\label{eq-8-4-abs-convergence}\tag{7.16} \end{equation}

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.

###### Example7.43
1. Explain why the series

\begin{equation*} 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} + \cdots \end{equation*}

must have a sum that is less than the series

\begin{equation*} \sum_{k=1}^{\infty} \frac{1}{k^2}\text{.} \end{equation*}
2. Explain why the series

\begin{equation*} 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} + \cdots \end{equation*}

must have a sum that is greater than the series

\begin{equation*} \sum_{k=1}^{\infty} -\frac{1}{k^2}\text{.} \end{equation*}
3. Given that the terms in the series

\begin{equation*} 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} + \cdots \end{equation*}

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

1. $1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \cdots \lt \sum_{k=1}^{\infty} \frac{1}{k^2} \text{.}$

2. $1 - \frac{1}{4} - \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} - \frac{1}{49} - \frac{1}{64} - \frac{1}{81} - \frac{1}{100} + \cdots \gt \sum_{k=1}^{\infty} -\frac{1}{k^2} \text{.}$

3. We expect this series to converge to some finite number between $\sum_{k=1}^{\infty} -\frac{1}{k^2}$ and $\sum_{k=1}^{\infty} \frac{1}{k^2}\text{.}$

Solution
1. Each term in the series

\begin{equation*} 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} + \cdots \end{equation*}

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

\begin{equation*} 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} + \cdots \lt \sum_{k=1}^{\infty} \frac{1}{k^2}\text{.} \end{equation*}
2. Each term in the series

\begin{equation*} 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} + \cdots \end{equation*}

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

\begin{equation*} 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} + \cdots \gt \sum_{k=1}^{\infty} -\frac{1}{k^2}\text{.} \end{equation*}
3. We know that $\sum_{k=1}^{\infty} \frac{1}{k^2}$ is a $p$-series with $p = 2 \gt 1$ and so is a convergent series. Since $\sum_{k=1}^{\infty} \frac{1}{k^2} = -\sum_{k=1}^{\infty} \frac{1}{k^2}\text{,}$ the series $\sum_{k=1}^{\infty} -\frac{1}{k^2}$ also converges. Since the terms in the series

\begin{equation*} 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} + \cdots \end{equation*}

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 $\sum_{k=1}^{\infty} \frac{1}{k^2}\text{.}$

As the example in Example7.43 suggests, if a series $\sum a_k$ has some negative terms but $\sum |a_k|$ converges, then the original series, $\sum a_k\text{,}$ must also converge. That is, if $\sum | a_k |$ converges, then so must $\sum a_k\text{.}$

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 $\sum \frac{1}{k^2}\text{.}$ But there are series, such as the alternating harmonic series $\sum (-1)^{k+1} \frac{1}{k}\text{,}$ that converge while the corresponding series of absolute values, $\sum \frac{1}{k}\text{,}$ diverges. We distinguish between these behaviors by introducing the following language.

###### Absolute vs. Conditional Convergence

Consider a series $\sum a_k\text{.}$

1. The series $\sum a_k$ converges absolutely (or is absolutely convergent) provided that $\sum | a_k |$ converges.

2. The series $\sum a_k$ converges conditionally (or is conditionally convergent) provided that $\sum | a_k |$ diverges and $\sum a_k$ converges.

In this terminology, the series(7.16) converges absolutely while the alternating harmonic series is conditionally convergent.

###### Example7.44
1. Consider the series $\sum (-1)^k \frac{\ln(k)}{k}\text{.}$

1. Does this series converge? Explain.

2. Does this series converge absolutely? Explain what test you use to determine your answer.

2. Consider the series $\sum (-1)^k \frac{\ln(k)}{k^2}\text{.}$

1. Does this series converge? Explain.

2. Does this series converge absolutely? Hint: Use the fact that $\ln(k) \lt \sqrt{k}$ for large values of $k$ and then compare to an appropriate $p$-series.

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

\begin{equation*} \lim_{k \to \infty} \frac{\ln(k)}{k} = \lim_{k \to \infty} \frac{1}{k} = 0\text{.} \end{equation*}

Also, $\frac{d}{dk} \frac{\ln(k)}{k} = \frac{1-\ln(k)}{k^2}$ is negative when $k \gt e\text{,}$ so the sequence $\left\{ \frac{\ln(k)}{k} \right\}$ 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{align*} \lim_{t \to \infty} \int_{1}^{t} \frac{\ln(x)}{x} \amp = \lim_{t \to \infty} \left. \frac{\ln(x)^2}{2} \right|_1^t\\ \amp = \lim_{t \to \infty} \frac{\ln(t)^2}{2}\\ \amp = \lim_{t \to \infty} \frac{\ln(t)^2}{2}\\ \amp = \infty\text{.} \end{align*}

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

\begin{equation*} \lim_{k \to \infty} \frac{\ln(k)}{k^2} = \lim_{k \to \infty} \frac{1}{2k^2} = 0\text{.} \end{equation*}

Also,

\begin{equation*} \frac{d}{dk} \frac{\ln(k)}{k^2} = \frac{k-2k\ln(k)}{k^4} = \frac{1-2\ln(k)}{k^3} \end{equation*}

is negative when $k \gt e\text{,}$ so the sequence $\left\{ \frac{\ln(k)}{k^2} \right\}$ 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

\begin{equation*} \lim_{k \to \infty} \frac{ \ln(k) }{ k^{1/2} } = \lim_{k \to \infty} \frac{2}{k^{1/2}} = 0\text{,} \end{equation*}

So $\frac{1}{\sqrt{k}}$ dominates $\ln(k)$ and $\ln(k) \lt \sqrt{k}$ for large $k\text{.}$ It follows that

\begin{equation*} \frac{\ln(k)}{k^2} \lt \frac{ \sqrt{k} }{k^2} = \frac{1}{k^{3/2}} \end{equation*}

for large $k\text{.}$ Therefore,

\begin{equation*} \sum \frac{\ln(k)}{k^2} \lt \sum \frac{1}{k^{3/2}} \end{equation*}

for large $k\text{.}$ Since $\sum \frac{1}{k^{3/2}}$ is a $p$-series with $p=\frac{3}{2} \gt 1\text{,}$ 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.

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.

### SubsectionSummary of Tests for Convergence of Series

###### Convergence Test Summary

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.

Geometric Series

The geometric series $\sum ar^k$ with ratio $r$ converges for $-1 \lt r \lt 1$ and diverges for $|r| \geq 1\text{.}$

The sum of the convergent geometric series $\displaystyle \sum_{k=0}^{\infty} ar^k$ is $\frac{a}{1-r}\text{.}$

Divergence Test

If the sequence $a_n$ does not converge to 0, then the series $\sum a_k$ diverges.

This is the first test to apply because the conclusion is simple. However, if $\lim_{n \to \infty} a_n = 0\text{,}$ no conclusion can be drawn.

Integral Test

Let $f$ be a positive, decreasing function on an interval $[c,\infty)$ and let $a_k = f(k)$ for each positive integer $k \geq c\text{.}$

• If $\int_c^{\infty} f(t) \ dt$ converges, then $\sum a_k$ converges.

• If $\int_c^{\infty} f(t) \ dt$ diverges, then $\sum a_k$ diverges.

Use this test when $f(x)$ is easy to integrate.

Direct Comp. Test

(see Ex 4 in Section7.3)

Let $0 \leq a_k \leq b_k$ for each positive integer $k\text{.}$

• If $\sum b_k$ converges, then $\sum a_k$ converges.

• If $\sum a_k$ diverges, then $\sum b_k$ diverges.

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

Let $a_n$ and $b_n$ be sequences of positive terms. If

\begin{equation*} \displaystyle \lim_{k \to \infty} \frac{a_k}{b_k} = L \end{equation*}

for some positive finite number $L\text{,}$ then the two series $\sum a_k$ and $\sum b_k$ either both converge or both diverge.

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

Let $a_k \neq 0$ for each $k$ and suppose

\begin{equation*} \displaystyle \lim_{k \to \infty} \frac{|a_{k+1}|}{|a_k|} = r\text{.} \end{equation*}
• If $r \lt 1\text{,}$ then the series $\sum a_k$ converges absolutely.

• If $r \gt 1\text{,}$ then the series $\sum a_k$ diverges.

• If $r=1\text{,}$ then test is inconclusive.

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

Root Test

(see Exercise 2 in Section7.3)

Let $a_k \geq 0$ for each $k$ and suppose

\begin{equation*} \displaystyle \lim_{k \to \infty} \sqrt[k]{a_k} = r\text{.} \end{equation*}
• If $r \lt 1\text{,}$ then the series $\sum a_k$ converges.

• If $r \gt 1\text{,}$ then the series $\sum a_k$ diverges.

• If $r=1\text{,}$ then test is inconclusive.

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

If $a_n$ is a positive, decreasing sequence so that $\displaystyle \lim_{n \to \infty} a_n = 0\text{,}$ then the alternating series $\sum (-1)^{k+1} a_k$ converges.

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.

Let $S_n = \displaystyle \sum_{k=1}^n (-1)^{k+1} a_k$ be the $n$th partial sum of the alternating series $\displaystyle \sum_{k=1}^{\infty} (-1)^{k+1} a_k\text{.}$ Assume $a_n \gt 0$ for each positive integer $n\text{,}$ the sequence $a_n$ decreases to 0 and $\displaystyle \lim_{n \to \infty} S_n = S\text{.}$ Then it follows that $|S - S_n| \lt a_{n+1}\text{.}$

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.

###### Example7.45

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.

1. $\displaystyle\sum_{k=3}^{\infty} \ \frac{2}{\sqrt{k-2}}$

2. $\displaystyle\sum_{k=1}^{\infty} \ \frac{k}{1+2k}$

3. $\displaystyle\sum_{k=0}^{\infty} \ \frac{2k^2+1}{k^3+k+1}$

4. $\displaystyle\sum_{k=0}^{\infty} \ \frac{100^k}{k!}$

5. $\displaystyle\sum_{k=1}^{\infty} \ \frac{2^k}{5^k}$

6. $\displaystyle\sum_{k=1}^{\infty} \ \frac{k^3-1}{k^5+1}$

7. $\displaystyle\sum_{k=2}^{\infty} \ \frac{3^{k-1}}{7^k}$

8. $\displaystyle\sum_{k=2}^{\infty} \ \frac{1}{k^k}$

9. $\displaystyle\sum_{k=1}^{\infty} \ \frac{(-1)^{k+1}}{\sqrt{k+1}}$

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

11. Determine a value of $n$ so that the $n$th partial sum $S_n$ of the alternating series $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln(n)}$ approximates the sum to within 0.001.

1. $\sum_{k=3}^{\infty} \ \frac{2}{\sqrt{k-2}}$ diverges.

2. $\sum_{k=1}^{\infty} \ \frac{k}{1+2k}$ diverges.

3. $\sum_{k=0}^{\infty} \ \frac{2k^2+1}{k^3+k+1}$ diverges.

4. $\sum_{k=0}^{\infty} \ \frac{100^k}{k!}$ converges.

5. $\sum_{k=1}^{\infty} \ \frac{2^k}{5^k}$ is a geometric series with ratio $\frac{2}{5}$ and sum $\frac{2}{3}\text{.}$

6. $\sum_{k=1}^{\infty} \ \frac{k^3-1}{k^5+1}$ converges.

7. $\sum_{k=2}^{\infty} \ \frac{3^{k-1}}{7^k}$ is a convergent geometric series with $a = \frac{3}{49}$ and $r = \frac{3}{7}$ and sum $\frac{3}{28}\text{.}$

8. $\sum_{k=2}^{\infty} \ \frac{1}{k^k}$ converges.

9. $\sum_{k=1}^{\infty} \ \frac{(-1)^{k+1}}{\sqrt{k+1}}$ converges.

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

11. $\sum_{k=2}^{\infty} \frac{(-1)^k}{\ln(k)}$ converges very slowly.

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

\begin{align*} \lim_{k \to \infty} \frac{\frac{2}{\sqrt{k-2}}}{\frac{2}{\sqrt{k}}} \amp = \lim_{k \to \infty} \frac{\sqrt{k}}{\sqrt{k-2}}\\ \amp = \lim_{k \to \infty} \sqrt{\frac{1}{1-\frac{2}{k}}}\\ \amp = 1\text{.} \end{align*}

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 $p=\frac{1}{2}\text{,}$ so $\sum_{k=3}^{\infty} \ \frac{2}{\sqrt{k}}$ diverges. Therefore, $\sum_{k=3}^{\infty} \ \frac{2}{\sqrt{k-2}}$ diverges as well.

2. Note that

\begin{equation*} \lim_{k \to \infty} \frac{k}{1+2k} = \lim_{k \to \infty} \frac{k}{2k} = \frac{1}{2}\text{,} \end{equation*}

so the Divergence Test shows that $\sum_{k=1}^{\infty} \ \frac{k}{1+2k}$ diverges.

3. For large values of $k\text{,}$ $\frac{2k^2+1}{k^3+k+1}$ looks like $\frac{2}{k}\text{.}$ We will use the Limit Comparison Test to compare these two series:

\begin{align*} \lim_{k \to \infty} \frac{\frac{2k^2+1}{k^3+k+1}}{\frac{2}{k}} \amp = \lim_{k \to \infty} \frac{(2k^2+1)k}{2(k^3+k+1)}\\ \amp = \lim_{k \to \infty} \frac{2 + \frac{1}{k^2}}{2+\frac{1}{k^2}+\frac{1}{k^3}}\\ \amp = 1\text{.} \end{align*}

Since this limit is a positive constant, we know that the two series $\sum_{k=1}^{\infty} \ \frac{2k^2+1}{k^3+k+1}$ 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 $p=1\text{,}$ so $\sum_{k=1}^{\infty} \ \frac{2}{k}$ diverges. Therefore, $\sum_{k=0}^{\infty} \ \frac{2k^2+1}{k^3+k+1}$ diverges as well.

4. Series that involve factorials are good candidates for the Ratio Test:

\begin{align*} \lim_{k \to \infty} \frac{\frac{100^{k+1}}{(k+1)!}}{\frac{100^k}{k!}} \amp = \lim_{k \to \infty} \frac{100^{k+1}k!}{100^k(k+1)!}\\ \amp = \lim_{k \to \infty} \frac{100}{k+1}\\ \amp = 0\text{.} \end{align*}

Since this limit is 0, the Ratio Test tells us that the series $\sum_{k=0}^{\infty} \ \frac{100^k}{k!}$ converges.

5. Notice that $\sum_{k=1}^{\infty} \ \frac{2^k}{5^k} = \sum_{k=1}^{\infty} \ \left(\frac{2}{5}\right)^k$ is a geometric series with ratio $\frac{2}{5}\text{.}$ Let's write out the first few terms to identify the value of $a\text{:}$

\begin{equation*} \sum_{k=1}^{\infty} \ \left(\frac{2}{5}\right)^k = \frac{2}{5} + \left(\frac{2}{5}\right)^2 + \left(\frac{2}{5}\right)^3 + \cdots = \left(\frac{2}{5}\right)\left[ 1 + \frac{2}{5} + \left(\frac{2}{5}\right)^2 + \cdots \right]\text{.} \end{equation*}

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

\begin{equation*} \sum_{k=1}^{\infty} \ \frac{2^k}{5^k} = \left(\frac{2}{5}\right) \left(\frac{1}{1-\frac{2}{5}} \right) = \left(\frac{2}{5}\right)\left(\frac{5}{3}\right) = \frac{2}{3}\text{.} \end{equation*}
6. For large values of $n\text{,}$ $\frac{k^3-1}{k^5+1}$ looks like $\frac{k^3}{k^5}=\frac{1}{k^2}\text{.}$ We will use the Limit Comparison Test to compare $\sum_{k=1}^{\infty} \ \frac{k^3-1}{k^5+1}$ and $\sum_{k=1}^{\infty} \ \frac{1}{k^2}\text{:}$

\begin{align*} \lim_{k \to \infty} \frac{\frac{k^3-1}{k^5+1}}{\frac{1}{k^2}} \amp = \lim_{k \to \infty} \frac{k^5-k^2}{k^5+1}\\ \amp = \lim_{k \to \infty} \frac{1-\frac{1}{k^3}}{1+\frac{1}{k^5}}\\ \amp = 1\text{.} \end{align*}

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 $p=2\text{,}$ 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.

7. This series looks geometric, but to be sure let's write out the first few terms of this series:

\begin{equation*} \sum_{k=2}^{\infty} \ \frac{3^{k-1}}{7^k} = \frac{3}{7^2} + \frac{3^2}{7^3} + \frac{3^3}{7^4} + \cdots = \left( \frac{3}{7^2} \right) \left[1 + \frac{3}{7} + \left( \frac{3}{7} \right)^2 + \left( \frac{3}{7} \right)^3 + \cdots \right]\text{.} \end{equation*}

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

\begin{align*} \sum_{k=2}^{\infty} \ \frac{3^{k-1}}{7^k} \amp = \left(\frac{3}{49}\right) \left(\frac{1}{1-\frac{3}{7}} \right)\\ \amp = \left(\frac{3}{49}\right)\left(\frac{7}{4}\right)\\ \amp = \frac{3}{28}\text{.} \end{align*}
8. $\sum_{k=2}^{\infty} \ \frac{1}{k^k}$ converges by direct comparison with the series $\sum_{k=2}^{\infty} \ \frac{1}{k^2}\text{.}$ Note that for all $k\geq 2\text{,}$ we have $k^k \geq k^2\text{,}$ so $\frac{1}{k^k}\leq \frac{1}{k^2}\text{.}$ Since $\sum_{k=2}^\infty \frac{1}{k^2}$ is a convergent $p$-series with $p=2\text{,}$ then by the direct comparison test, $\sum_{k=2}^{\infty} \frac{1}{k^k}$ also converges.

9. This series is an alternating series of the form $\sum_{k=1}^{\infty} (-1)^{k+1} a_k$ with $a_k = \frac{1}{\sqrt{k+1}}\text{.}$ Since

\begin{equation*} \lim_{k \to \infty} \frac{1}{\sqrt{k+1}} = 0 \end{equation*}

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.

10. For this example we will use the Integral Test with the substitution $w = \ln(x)\text{,}$ $dw = \frac{1}{x} \ dx\text{:}$

\begin{align*} \displaystyle \int_2^{\infty} \frac{1}{x \ln(x)} \ dx \amp = \lim_{b \to \infty} \int_2^b \frac{1}{x \ln(x)} \ dx\\ \amp = \displaystyle \lim_{b \to \infty} \int_{\ln(2)}^{\ln(b)} \frac{1}{w} \ dw\\ \amp = \displaystyle \lim_{b \to \infty} \ln(w) \bigm|_{\ln(2)}^{\ln(b)}\\ \amp = \displaystyle \lim_{b \to \infty} \left[ \ln(\ln(b)) - \ln(\ln(2)) \right]\\ \amp = \infty\text{.} \end{align*}

Since $\int_2^{\infty} \frac{1}{x \ln(x)} \ dx$ diverges, we conclude $\sum_{k=2}^{\infty} \ \frac{1}{k \ln(k)}$ diverges.

11. First note that the terms $\frac{1}{\ln(k)}$ decrease to $0\text{,}$ 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

\begin{equation*} |S_n - S| \lt |S_n - S_{n+1}| = a_{n+1} \end{equation*}

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

\begin{align*} a_{n+1} \amp \lt 0.001\\ \frac{1}{\ln(n+1)} \amp \lt 0.001\\ \ln(n+1) \amp \gt 1000\\ n+1 \amp \gt e^{1000}\\ n \amp \gt e^{1000}-1\text{.} \end{align*}

So if we choose $n = e^{1000}\text{,}$ 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.

### SubsectionSummary

• An alternating series is a series whose terms alternate in sign. It has the form

\begin{equation*} \sum (-1)^ka_k \end{equation*}

where $a_k$ is a positive real number for each $k\text{.}$

• The sequence of partial sums of a convergent alternating series oscillates around the sum of the series if the sequence of $n$th terms converges to 0. That is why the Alternating Series Test shows that the alternating series $\sum_{k=1}^{\infty} (-1)^ka_k$ converges whenever the sequence $\{a_n\}$ of $n$th terms decreases to 0.

• 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 $\sum_{k=1}^{\infty} (-1)^ka_k$ is $|S_n - S_{n-1}| = a_n\text{.}$ Since the partial sums oscillate around the sum $S$ of the series, it follows that

\begin{equation*} |S - S_n| \lt a_n\text{.} \end{equation*}

So the $n$th partial sum of a convergent alternating series $\sum_{k=1}^{\infty} (-1)^ka_k$ approximates the actual sum of the series to within $a_n\text{.}$

### SubsectionExercises

Conditionally convergent series converge very slowly. As an example, consider the famous formula8We will derive this formula in upcoming work.

\begin{equation} \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots = \sum_{k=0}^{\infty} (-1)^{k} \frac{1}{2k+1}\text{.}\label{Ex-8-4-pi}\tag{7.17} \end{equation}

In theory, the partial sums of this series could be used to approximate $\pi\text{.}$

1. Show that the series in (7.17) converges conditionally.

2. Let $S_n$ be the $n$th partial sum of the series in (7.17). Calculate the error in approximating $\frac{\pi}{4}$ with $S_{100}$ and explain why this is not a very good approximation.

3. Determine the number of terms it would take in the series (7.17) to approximate $\frac{\pi}{4}$ 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.)

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 $S_n\text{.}$ This suggests that the average $\frac{S_n+S_{n+1}}{2}$ is a better approximation to $S$ than is $S_n\text{.}$

1. Show that $\frac{S_n+S_{n+1}}{2} = S_n + \frac{1}{2}(-1)^{n+2} a_{n+1}\text{.}$

2. Use this revised approximation in (a) with $n = 20$ to approximate $\ln(2)$ given that

\begin{equation*} \ln(2) = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k}\text{.} \end{equation*}

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

In this exercise, we examine one of the conditions of the Alternating Series Test. Consider the alternating series

\begin{equation*} 1 - 1 + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{9} + \frac{1}{4} - \frac{1}{16} + \cdots\text{,} \end{equation*}

where the terms are selected alternately from the sequences $\left\{\frac{1}{n}\right\}$ and $\left\{-\frac{1}{n^2}\right\}\text{.}$

1. Explain why the $n$th term of the given series converges to 0 as $n$ goes to infinity.

2. Rewrite the given series by grouping terms in the following manner:

\begin{equation*} (1 - 1) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{9}\right) + \left(\frac{1}{4} - \frac{1}{16}\right) + \cdots\text{.} \end{equation*}

Use this regrouping to determine if the series converges or diverges.

3. Explain why the condition that the sequence $\{a_n\}$ decreases to a limit of 0 is included in the Alternating Series Test.

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

\begin{equation*} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = \ln(2)\text{,} \end{equation*}

a fact which will be verified in a later section.

1. First we make a quick analysis of the positive and negative terms of the alternating harmonic series.

1. Show that the series $\sum_{k=1}^{\infty} \frac{1}{2k}$ diverges.

2. Show that the series $\sum_{k=1}^{\infty} \frac{1}{2k+1}$ diverges.

3. Based on the results of the previous parts of this exercise, what can we say about the sums $\sum_{k=C}^{\infty} \frac{1}{2k}$ and $\sum_{k=C}^{\infty} \frac{1}{2k+1}$ for any positive integer $C\text{?}$ Be specific in your explanation.

2. Recall addition of real numbers is commutative; that is

\begin{equation*} a + b = b + a \end{equation*}

for any real numbers $a$ and $b\text{.}$ 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 $S\text{,}$ we can rearrange the terms of the alternating harmonic series $\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}$ so that the sum is $S\text{.}$ 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 $S\text{.}$

1. Explain how we know that, regardless of the value of $S\text{,}$ we can find a partial sum $P_1$

\begin{equation*} P_1 = \sum_{k=1}^{n_1} \frac{1}{2k+1} = 1 + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{2n_1+1} \end{equation*}

of the positive terms of the alternating harmonic series that equals or exceeds $S\text{.}$ Let

\begin{equation*} S_1 = P_1\text{.} \end{equation*}
2. Explain how we know that, regardless of the value of $S_1\text{,}$ we can find a partial sum $N_1$

\begin{equation*} N_1 = -\sum_{k=1}^{m_1} \frac{1}{2k} = -\frac{1}{2} - \frac{1}{4} - \frac{1}{6} - \cdots - \frac{1}{2m_1} \end{equation*}

so that

\begin{equation*} S_2 = S_1 + N_1 \leq S\text{.} \end{equation*}
3. Explain how we know that, regardless of the value of $S_2\text{,}$ we can find a partial sum $P_2$

\begin{equation*} P_2 = \sum_{k=n_1+1}^{n_2} \frac{1}{2k+1} = \frac{1}{2(n_1+1)+1} + \frac{1}{2(n_1+2)+1} + \cdots + \frac{1}{2n_2+1} \end{equation*}

of the remaining positive terms of the alternating harmonic series so that

\begin{equation*} S_3 = S_2 + P_2 \geq S\text{.} \end{equation*}
4. Explain how we know that, regardless of the value of $S_3\text{,}$ we can find a partial sum

\begin{equation*} N_2 = -\sum_{k=m_1+1}^{m_2} \frac{1}{2k} = -\frac{1}{2(m_1+1)} - \frac{1}{2(m_1+2)} - \cdots - \frac{1}{2m_2} \end{equation*}

of the remaining negative terms of the alternating harmonic series so that

\begin{equation*} S_4 = S_3 + N_2 \leq S\text{.} \end{equation*}
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 $\lim_{n \to \infty} S_n = S\text{.}$