Question

Use the ratio test for absolute convergence to determine whether the series in Exercises 30–35 converge absolutely or diverge.
$\sum _{k=0}^{\infty }\frac{{(-3)}^{k+1}}{\left(2k\right)!}$

Short answer

The series $\sum _{k=0}^{\infty }\frac{{(-3)}^{k+1}}{\left(2k\right)!}$converges absolutely.

Step-by-step solution

Step 1. Given Information.

The series:

$\sum _{k=0}^{\infty }\frac{{(-3)}^{k+1}}{\left(2k\right)!}$

Step 2. Rewrite the series.

$a_k=\frac{{(-3)}^{k+1}}{\left(2k\right)!}$

Step 3. Find ak+1.

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

Step 4. Calculate ak+1ak.

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

Step 5. Take limits.

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

So by the ratio test, the series converges absolutely.