Question

Use any convergence tests to determine whether the series converge absolutely, converge conditionally, or diverge. Explain why the series meets the hypotheses of the test you select.

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

Short answer

The series is absolutely convergent.

Step-by-step solution

Step 1. Given information.

Consider the given question,

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

Step 2. Consider the ratio.

The general term of the series $\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }{\left(-1\right)}^k{e}^{-k}$is $a_k={\left(-1\right)}^k{e}^{-k}$.

The ratio $\left|\frac{a_{k+1}}{a_k}\right|$gives,

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

Step 3. Find the value of the limit.

The value of $\underset{k\to \infty }{\lim }\left|\frac{a_{k+1}}{a_k}\right|$is given below,

$$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{a_{k+1}}{a_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{1}{e}\right| \\ =\frac{1}{e} \end{gathered}$$

The value of the above limit is less than $1$.

Thus, by ratio test for absolute convergence, the series $\sum _{k=1}^{\infty }{\left(-1\right)}^k{e}^{-k}$is absolutely convergent.

Hence, the series is absolutely convergent.