Answer
Converges absolutely
Work Step by Step
The series will converge conditionally when it is a convergent series, but the condition is when the series of its absolute value diverges and the series will converge absolutely but the condition is when the series of its absolute value converges.
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, the absolute value of the given series is : $a_k=k e^{-k}$
$\lim\limits_{k \to \infty} a_k= \lim\limits_{k \to \infty}\dfrac{(k+1) e^{-k}}{k e^{-k}}=\dfrac{1}{e} \lim\limits_{k \to \infty}(1+\dfrac{1}{k})$
Since, $n \gt 0$ for all the values of $n$ and $\lim\limits_{x \to \infty} x^n=\infty$ and $\lim\limits_{x \to \infty} x^{-n}=0$
$\lim\limits_{k \to \infty} a_k= \dfrac{1}{e} \times (1+0)=\dfrac{1}{e} \lt 1$
This implies that the absolute series $\Sigma_{k=1}^{\infty} ke^{-k}$ converges, so the given series converges absolutely.
