Question

Use the ratio test for absolute convergence to determine whether the series in Exercises 30–35 converge absolutely or diverge.

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

Short answer

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

Step-by-step solution

Step 1. Given Information.

The series:

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

Step 2. Rewrite the series.

$a_k=\sum _{k=1}^{\infty }\frac{{(-k)}^k}{k!}$

Step 3. Find ak+1.

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

Step 4. Calculate ak+1ak.

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

Step 5. Take limits.

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

So by the ratio test, the series converges absolutely.