Answer
$\displaystyle\sum_{k=1}^{\infty}ke^{-k}$ is convergent.
Work Step by Step
\[\sum_{k=1}^{\infty}ke^{-k}=\sum_{n=1}^{\infty}ne^{-n}\]
Let \[\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}ne^{-n}\]
\[\Rightarrow a_n=ne^{-n}\]
So $a_n$ is positive term and montonically decreasing.
So Integral test is applicable.
Let \[I=\int_{1}^{\infty}xe^{-x}dx\]
\[I=\lim_{t\rightarrow \infty}\int_{1}^{t}xe^{-x}dx\;\;\;\;\;\;\ldots (1)\]
Let \[I_1=\int xe^{-x}dx\]
Using integration by parts:
\[I_1=-xe^{-x}+\int e^{-x}dx\]
\[I_1=-xe^{-x}-e^{-x}=-e^{-x}(x+1)=\frac{-(x+1)}{e^x}\]
\[I_1=\frac{-(x+1)}{e^x}\;\;\;\;\;\;\;\;\;\ldots (2)\]
Using (2) in (1)
\[I=\lim_{t\rightarrow \infty}\left[\frac{-(x+1)}{e^x}\right]_{1}^{t}\]
\[\Rightarrow I=\lim_{t\rightarrow \infty}\left[-\frac{(t+1)}{e^t}+\frac{2}{e}\right]\]
\[\Rightarrow I=-\lim_{t\rightarrow \infty}\left[\frac{(t+1)}{e^t}\right]+\frac{2}{e}\]
Which is $\displaystyle\frac{\infty}{\infty}$ case, using L' Hopitals rule
\[\Rightarrow I=-\lim_{t\rightarrow \infty}\left[\frac{1}{e^t}\right]+\frac{2}{e}\]
\[\Rightarrow I=0+\frac{2}{e}=\frac{2}{e}\]
Which is finite so $I$ is convergent.
So by Integral test, $\displaystyle\sum_{k=1}^{\infty}ke^{-k}$ is convergent.
