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.

$\sum _{k=1}^{\infty }{\left(-1\right)}^k{e}^{-k^2}$

Short answer

The series converges absolutely.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\sum _{k=1}^{\infty }{\left(-1\right)}^k{e}^{-k^2}$

Step 2. Consider the general series.

The general term of the series $\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }{\left(-1\right)}^k{e}^{-k^2}$is given by,

$a_k={\left(-1\right)}^k{e}^{-k^2}$

The ratio role="math" localid="1649154069574" $\left|\frac{a_{k+1}}{a_k}\right|$is given by,

role="math" localid="1649154169585" $$\begin{gathered} \left|\frac{a_{k+1}}{a_k}\right|=\left|\frac{{\left(-1\right)}^{k+1}{e}^{-{\left(k+1\right)}^2}}{{\left(-1\right)}^k{e}^{-k^2}}\right| \\ =\left|\frac{\left(-1\right)e^{-{\left(k+1\right)}^2}}{e^{-k^2}}\right| \\ =\left|\frac{e^{k^2}}{e^{{(k+1)}^2}}\right| \end{gathered}$$

Step 3. Find the value of limit.

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

role="math" localid="1649154467758" $$\begin{gathered} \underset{k\to \infty }{\lim }\left|\frac{a_{k+1}}{a_k}\right|=\underset{k\to \infty }{\lim }\left|\frac{e^{k^2}}{e^{{\left(k+1\right)}^2}}\right| \\ =\underset{k\to \infty }{\lim }\left|\frac{1}{e^{2k+1}}\right| \\ =\underset{k\to \infty }{\lim }\left|\frac{1}{e·e^{2k}}\right| \\ =0 \end{gathered}$$

The value of $\underset{k\to \infty }{\lim }\left|\frac{a_{k+1}}{a_k}\right|$is $0$, which is less than $1$.

Thus, by ratio test for absolute convergence, the series $\sum _{k=1}^{\infty }{\left(-1\right)}^k{e}^{-k^2}$ is absolutely convergent.

Hence, the given series is absolutely convergent.