Question

Conditional and absolute convergence: For each of the series that follow, determine whether the series converges absolutely, converges conditionally, or diverges. Explain the criteria you are using and why your conclusion is valid.

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

Short answer

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

Step-by-step solution

Step 1. Given Information.

The series:

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

Step 2. By Alternating Series Test.

According to the Alternating Series Test, the sequence $a_{k+1}<a_k$for every . Then the alternating series $a_{k+1},a_k$both converges.

Step 3. Find ak+1.

$$\begin{gathered} a_k=e^{-k} \\ {a}_{k+1}=e^{-(k+1)} \\ =e^{-k-1} \\ {a}_{k+1}<a_k \end{gathered}$$

So the sequence is monotonic decreasing sequence.

Step 4. Find limk→∞ak.

$$\begin{gathered} \underset{k\to \infty }{lim}{a}_k=\underset{k\to \infty }{lim}{e}^{-k} \\ =0 \end{gathered}$$

So the series converges.

Step 5. Find ak+1ak.

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

So the series converges absolutely.