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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The series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{2}}$ is a convergent p-series ($p > 1).$
So we take $b_{n}= \displaystyle \frac{1}{n^{2}}$
Taking $a_{n}=\displaystyle \frac{1}{n!}$ and using the discussion in Example 5 of Sec 9-1,
for large n,
$n! > 2^{n}$, and
the exponential function $2^{x}$ grows much faster than $x^{2}$, so , for large n, $(n \geq 5)$
$n! > 2^{n} > n^{2}$ , so
$0 < \displaystyle \frac{1}{n!} < \frac{1}{n^{2}}$
Since $0 < a_{n} \leq b_{n} ,$ case 1 applies,
that is, $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n!}$ also converges.
