Question

Do the following to $\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)!}$:

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${10}^{-6}$ofL.

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$\left[0.5403,0.5403\right]$.

Part (c): The smallest value of n is $5$.

Step-by-step solution

Part (a) Step 1. Given information.

Consider the given question,

$\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)!}$

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

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

Also, $\underset{k\to \infty }{\lim }\frac{1}{\left(2k\right)!}=0$

Thus, the conditions of alternating series test are fulfilled.

Hence, the alternating series$\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k}{\left(2k\right)!}$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,

$$\begin{gathered} S_{10}=1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+...-\frac{1}{18!} \\ =-0.5403 \end{gathered}$$

Thus, the value of$S_{10}$is$-0.5403$.

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 $\frac{1}{22}=8.89\times {10}^{-22}$. That is given below,

$\left|L-0.5403\right|<4.68\times {10}^{-23}$

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

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

Then, the interval is$\left[S_{10}-a_{11},S_{10}\right]=\left[0.5403,0.5403\right]$.

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

The smallest value of n is given below,

$$\begin{gathered} \frac{1}{\left(2n\right)!}\le {10}^{-6} \\ {10}^6\le \left(2n\right)! \\ 5\le n \end{gathered}$$

Thus, the smallest value of n is$5$.