Question

In Exercises 52–57, do each of the following:

(a) Show that the given alternating series converges.

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

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

$$\sum _{k=0}^{\mathrm{\infty }}{\displaystyle \frac{(-1{)}^k}{(2k+1)!}}$$

Short answer

(a). Using the test for convergence for alternating series we can see that the given alternating series converges.

(b). The value of $$S_{10}$$ is $$1-{\displaystyle \frac{1}{3!}}+{\displaystyle \frac{1}{5!}}-{\displaystyle \frac{1}{7!}}+{\displaystyle \frac{1}{9!}}-{\displaystyle \frac{1}{11!}}+{\displaystyle \frac{1}{13!}}-{\displaystyle \frac{1}{15!}}+{\displaystyle \frac{1}{17!}}-{\displaystyle \frac{1}{19!}}+{\displaystyle \frac{1}{21!}}\approx 0.841470984$$

and the value of $$L$$ is $$L\in (0.84147098480789650665254,0.8414709848078965066526)$$.

Step-by-step solution

Step 1. Given Information

Given, an alternating series

$$\sum _{k=0}^{\mathrm{\infty }}{\displaystyle \frac{(-1{)}^k}{(2k+1)!}}$$

Step 2. Using the alternating series test

Alternating series test for convergence is stated as if $$\{a_k\}$$ be a strictly decreasing sequence of positive numbers such that $$\underset{k\to \mathrm{\infty }}{lim}{a}_k=0$$ . Then the alternating series $$\sum _{k=1}^{\mathrm{\infty }}(-1{)}^k{a}_k$$ and $$\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}{a}_k$$ both converges.

Step 3. Applying the alternating series test to the given series 

The series $$\sum _{k=0}^{\mathrm{\infty }}{\displaystyle \frac{1}{(2k+1)!}}$$ is clearly a nonnegative strictly decreasing series. Also,

$$\underset{k\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{(2k+1)!}}=0$$. Hence, by alternating series test this series converges and the series $$\sum _{k=0}^{\mathrm{\infty }}{\displaystyle \frac{(-1{)}^k}{(2k+1)!}}$$ also converges.

Step 4. Given information

We are given with Theorem 7.8 to approximate the partial sum $$S_{10}$$

Step 5. Calculating $$S_{10}$$

If we add the first 10 terms of the series to obtain the 10th partial sum, $$S_{10}$$, we obtain

$$S_{10}=1-{\displaystyle \frac{1}{3!}}+{\displaystyle \frac{1}{5!}}-{\displaystyle \frac{1}{7!}}+{\displaystyle \frac{1}{9!}}-{\displaystyle \frac{1}{11!}}+{\displaystyle \frac{1}{13!}}-{\displaystyle \frac{1}{15!}}+{\displaystyle \frac{1}{17!}}-{\displaystyle \frac{1}{19!}}+{\displaystyle \frac{1}{21!}}\approx 0.841470984$$

This is an approximation for the sum L of the series. By Theorem 7.38, the error in using $$S_{10}$$ is at most $${\displaystyle \frac{1}{23!}}\approx 3.86817\times {10}^{-23}$$. In addition, Theorem 7.38 tells us that $$S_{10}$$ is less than the sum of the series, $$L$$, because $$L-S_{10}$$has the same sign as the term $${\displaystyle \frac{1}{23!}}\approx 3.86817\times {10}^{-23}$$ . Also,

$$L-S_{10}<a_{11}\approx 3.86817\times {10}^{-23}$$. We have, $$L\in (S_{10},S_{10}+a_{11})$$. Therefore, $$L\in (0.84147098480789650665254,0.8414709848078965066526)$$

Step 6. Determining $$n$$

We have to find $$n$$ such that

$$S_n<{10}^{-6}$$