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 $$L$$.

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

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 $$1 - \dfrac{1}{3!} + \dfrac{1}{5!} -\dfrac{1}{7!} + \dfrac{1}{9!} -\dfrac{1}{11!} + \dfrac{1}{13!} - \dfrac{1}{15!} + \dfrac{1}{17!} - \dfrac{1}{19!} + \dfrac{1}{21!} \approx 0.841470984$$

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

Step-by-step solution

Step 1. Given Information

Given, an alternating series

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

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 \[ \lim_{k\to\infty} a_{k}=0 \] . Then the alternating series \[ \sum_{k=1}^{\infty} (-1)^{k} a_{k} \] and \[ \sum_{k=1}^{\infty} (-1)^{k+1} a_{k} \] both converges.

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

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

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

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

This is an approximation for the sum L of the series. By Theorem 7.38, the error in using $$S_{10}$$ is at most $$\dfrac{1}{23!} \approx 3.86817 \times 10^{-23}$$. 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 $$\dfrac{1}{23!} \approx 3.86817 \times 10^{-23}$$ . 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

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