Answer
Converges by Th. 9.12.
Work Step by Step
Th. 9.12 Direct Comparison Test:
Let $0 < a_{n} \leq b_{n}$ for all $n$.
1. If $\displaystyle \sum_{n=1}^{\infty}b_{n}$ converges, then $\displaystyle \sum_{n=1}^{\infty}a_{n}$ converges.
2. If $\displaystyle \sum_{n=1}^{\infty}a_{n}$ diverges, then $\displaystyle \sum_{n=1}^{\infty}b_{n}$ diverges.
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$\displaystyle \sum_{n=0}^{\infty}(\frac{1}{e})^{n}$ is a convergent geometric series $(r=\displaystyle \frac{1}{e} < 1)$
... so we aim for case (1).
Let $b_{n}=\displaystyle \frac{1}{e^{n}}$. Then
$a_{n}=\displaystyle \frac{1}{e^{n^{2}}} \leq \frac{1}{e^{n}} $, that is
$0 < a_{n} \leq b_{n}$ , so
Since $\displaystyle \sum_{n=1}^{\infty}b_{n}$ converges,
$\displaystyle \sum_{n=1}^{\infty}a_{n}=\sum_{n=0}^{\infty}\frac{1}{e^{n^{2}}}$ also converges
