Answer
Converges absolutely
Work Step by Step
Let us consider a Ratio Test for a series $l=\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_n}|$
1) When $l \lt 1$. then series $\Sigma a_n$ converges absolutely.
2) When $l \gt 1$. then series $\Sigma a_n$ diverges.
3) When $l = 1$. then series $\Sigma a_n$ may converge or diverge, that is, the test is inconclusive.
Here, we have $a_k=\dfrac{e^k}{(k+1)!}$
Now, $l=\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_n}|=\lim\limits_{n \to \infty}|\dfrac{\dfrac{e^{k+1}}{(k+2)(k+1)!}}{\dfrac{e^k}{(k+1)!}}|$
or, $l=\lim\limits_{k \to \infty}\dfrac{e^k}{(k+1)}$
Since, $n \gt 0$ for all the values of $n$ and $\lim\limits_{x \to \infty} x^n=\infty$ and $\lim\limits_{x \to \infty} x^{-n}=0$
Thus. $l=0 \lt 1$
So. the series $\Sigma_{k=1}^\infty \dfrac{e^k}{(k+1)!}$ converges by the ratio test. Thus, the given series $\Sigma_{k=1}^\infty \dfrac{(-1)^{k+1} e^k}{(k+1)!}$ converges absolutely.
