Answer
$\displaystyle\sum_{n=1}^{\infty}ne^{-n^2}$ is convergent
Work Step by Step
\[\sum_{k=1}^{\infty}ke^{-k^2}=\sum_{n=1}^{\infty}ne^{-n^2}\]
Let \[\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}ne^{-n^2}\]
\[\Rightarrow a_n=ne^{-n^2}=\frac{n}{e^{n^2}}\]
$a_n$ is positive term and monotonically decreasing.
so integral test is applicable.
Let \[I=\int_{1}^{\infty}xe^{-x^2}dx\]
\[\Rightarrow I=\lim_{t\rightarrow \infty}\int_{1}^{t} xe^{-x^2}dx\;\;\;\;\;\;\;\ldots (1)\]
Let \[I_1=\int xe^{-x^2}dx\]
Substitute $y=-x^2\;\;\Rightarrow\;\;dy=-2xdx\;\Rightarrow \displaystyle\frac{-1}{2}dy=xdx$
\[\Rightarrow I_1=\frac{-1}{2}\int e^ydy=\frac{-1}{2}e^y\]
\[\Rightarrow I_1=\frac{-1}{2}e^{-x^2}=\frac{-1}{2e^{x^2}}\;\;\;\;\;\;\;\ldots (2)\]
Using (2) in (1)
\[\Rightarrow I=\lim_{t\rightarrow \infty}\left[\frac{-1}{2e^{x^2}}\right]_{1}^{t}\]
\[\Rightarrow I=\lim_{t\rightarrow \infty}\left[\frac{-1}{2e^{t^2}}+\frac{1}{2e}\right]\]
\[\Rightarrow I=0+\frac{1}{2e}=\frac{1}{2e}\]
Which is finite so $I$ is convergent.
By Integral test, $\displaystyle\sum_{n=1}^{\infty}ne^{-n^2}$ is convergent.
