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 $$L$$.

\[ \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1} k!}{(2k+1)!} \]

Given, an alternating series

\[ \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1} k!}{(2k+1)!} \]

Alternating series test for convergence is stated as if $$\{ a_{k} \}$$ be a strictly decreasing sequence of positive numbers such that \[ \lim_{k\to\infty} a_{k}=0 \] . Then the alternating series \[ \sum_{k=1}^{\infty} (-1)^{k} a_{k} \] and \[ \sum_{k=1}^{\infty} (-1)^{k+1} a_{k} \] both converges.

The series \[ \sum_{k=0}^{\infty} \dfrac{k!}{(2k+1)!} \] is clearly a nonnegative strictly decreasing series. Also,

\[ \lim_{k\to\infty} \dfrac{k!}{(2k+1)!} =0 \]. Hence, by alternating series test this series converges and the series \[ \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1} k!}{(2k+1)!} \] also converges.

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

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

$$S_{10} = -1 + \dfrac{1}{3!} -\dfrac{2!}{5!} +\dfrac{3!}{7!} - \dfrac{4!}{9!} +\dfrac{5!}{11!} -\dfrac{6!}{13!} + \dfrac{7!}{15!} - \dfrac{8!}{17!} + \dfrac{9!}{19!} - \dfrac{10!}{21!} \approx -0.848872767004046$$

This is an approximation for the sum L of the series. By Theorem 7.38, the error in using $$S_{10}$$ is at most $$\dfrac{11!}{23!} \approx 1.54405 \times 10^{-15}$$. 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 $$\dfrac{11!}{23!} \approx 1.54405 \times 10^{-15}$$ . Also,

$$L-S_{10} < a_{11} \approx 1.54405 \times 10^{-15}$$. We have, $$L \in ( S_{10}, S_{10}+ a_{11})$$. Therefore, $$L \in (-0.848872767004046 , -0.848872766849641)$$

We have to find $$n$$ such that

$$S_{n} < 10^{-6}$$