Answer
The series converges.
Work Step by Step
We begin with the series:
$$\sum_{k=1}^{\infty}ke^{-k}$$
Using the ratio test, we know that the series converges if the following limit converges:
$$\lim_{k\to\infty}\frac{(k+1)e^{-(k+1)}}{ke^{-k}}$$
Algebraically manipulating the limit, we get:
$$\lim_{k\to\infty}\frac{(k+1)e^{-(k+1)}}{ke^{-k}}\\
=\lim_{k\to\infty}\frac{(k+1)e^{-k-1}}{ke^{-k}}\\
=\lim_{k\to\infty}\frac{1}{e}\frac{(k+1)e^{-k}}{ke^{-k}}\\
=\lim_{k\to\infty}\frac{1}{e}\frac{k+1}{k}\\
=\frac{1}{e}$$
Since $\frac{1}{e}<1$, the series converges.
