Question

Use the ratio test for absolute convergence to determine whether the series in Exercises 30–35 converge absolutely or diverge.
$\sum _{k=0}^{\infty }\frac{{(-7)}^k}{(2k+1)!}$

Short answer

The series $\sum _{k=0}^{\infty }\frac{{(-7)}^k}{(2k+1)!}$converges absolutely.

Step-by-step solution

Step 1. Given Information.

The series:

$\sum _{k=0}^{\infty }\frac{{(-7)}^k}{(2k+1)!}$

Step 2. Rewrite the series.

$a_k=\frac{{(-7)}^k}{(2k+1)!}$

Step 3. Find ak+1.

$$\begin{gathered} a_{k+1}=\frac{{(-7)}^{k+1}}{\left(2\right(k+1)+1)!} \\ =\frac{{(-7)}^{k+1}}{(2k+2+1)!} \\ =\frac{{(-7)}^{k+1}}{(2k+3)!} \end{gathered}$$

Step 4. Calculate ak+1ak.

$$\begin{gathered} \left|\frac{a_{k+1}}{a_k}\right|=\left|\frac{\frac{{(-7)}^{k+1}}{(2k+3)!}}{\frac{{(-7)}^k}{(2k+1)!}}\right| \\ =\left|\frac{{(-7)}^{k+1}(2k+1)!}{{(-7)}^k(2k+3)!}\right| \\ =\left|\frac{-7{(-7)}^k(2k+1)!}{{(-7)}^k(2k+3)(2k+2)(2k+1)!}\right| \\ =\left|\frac{-7}{(2k+3)(2k+2)}\right| \\ =\frac{7}{2(2k+3)(k+1)} \end{gathered}$$

Step 5. Take limits.

$$\begin{gathered} \underset{k\to \infty }{lim}\left|\frac{a_{k+1}}{a_k}\right|=\underset{k\to \infty }{lim}\frac{7}{2(2k+3)(k+1)} \\ =0 \end{gathered}$$

So by the ratio test, the series converges absolutely.