Question

Do the following to $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k}{k^2}$:

Part (a): Show that the given alternating series converges.

Part (b): Compute $S_{10}$and use Theorem $7.38$to find an interval containing the sum L of the series.

Part (c): Find the smallest value of n such that Theorem $7.38$guarantees that $S_n$is within ${10}^{-6}$ of L.

Short answer

Part (a): To show that the given alternating series converges, use the conditions of alternating series test.

Part (b): The interval containing the sum L of the series is $\left[0.08179,0.08262\right]$.

Part (c): The smallest value of n is$999$.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given question,

$\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k}{k^2}$

Part (a) Step 2. To show the alternating series converges.

The sequence $\left\{a_k\right\}=\left\{\frac{1}{k^2}\right\}$is a sequence of positive numbers and is a decreasing sequence as $\frac{1}{{\left(k+1\right)}^2}<\frac{1}{k^2}$.

Also, $\underset{k\to \infty }{\lim }\frac{1}{k^2}=0$

Thus, the conditions of alternating series test are fulfilled.

Hence, the alternating series $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k}{k^2}$ is convergent.

Part (b) Step 1. Find the interval containing the sum L of the series.

The value $S_{10}$is obtained by adding first $10$terms of the series to obtain the ${10}^{th}$partial sum. Therefore,$$\begin{gathered} S_{10}=1-\frac{1}{2^2}+\frac{1}{3^2}-...-\frac{1}{{10}^2} \\ =0.8179 \end{gathered}$$

Thus, the value of$S_{10}$is$0.8179$.

Part (b) Step 2.  Assume L to be the sum of an alternating series.

Assume L to be the sum of an alternating series satisfying the hypothesis of the alternating series test.

For any term $S_n$is the sequence of partial sums, $\left|L-S_n\right|<a_{n+1}$

Furthermore, the sign of the difference $L-S_n$is the sign of the coefficient of the term $a_{n+1}$.

The error in using $S_{10}$is at almost $\frac{1}{{11}^2}=0.0083$. That is given below,

$\left|L-0.8179\right|<0.0083$

The approximation $S_{10}$ is less than L as the coefficient of $a_{11}$is positive.

Thus, actual value of L lies in the interval $\left[S_{10},S_{10}+a_{11}\right]$.

Then, the interval is $$\begin{gathered} \left[S_{10},S_{10}+a_{11}\right]=\left[0.08179,0.08179+0.0083\right] \\ =\left[0.08179,0.08262\right] \end{gathered}$$

Part (c) Step 1. Find the smallest value of n.

The smallest value of n is given below,

$$\begin{gathered} \frac{1}{{\left(n+1\right)}^2}\le {10}^{-6} \\ {\left(n+1\right)}^{-2}\le {10}^{-6} \\ 1000\le n+1 \\ 999\le n \end{gathered}$$

Thus, the smallest value of n is$999$.