<s>[INST] Find the convergence of the series $\sum \frac{n^{n-2}}{e^n n!}$ Show that the series $\sum_{n=1}^{\infty} \frac{n^{n-2}}{e^n n!}$ is convergent.
I tried to use root test but it yields 1 which makes the test indecisive . Any other approach ?
 [/INST] As @robjohn pointed out in the comments theorem 4 from this answer suggest
$$
1 + \frac{1}{12\left(n + \frac{1}{2}\right)}
\le \frac{n! e^n}{n^n \sqrt{2 \pi n}}
\le 1 + \frac{1}{12\left(n - \frac{1}{3}\right)}.
$$
Simplifying the sums gives
$$
\frac{12n + 7}{12n + 6}
\le \frac{n! e^n}{n^n \sqrt{2 \pi n}}
\le \frac{12n - 3}{12n - 4}
$$
Now use $a \le b \le c \iff \frac{1}{c} \le \frac{1}{b} \le \frac{1}{a}$ and divide every term by $n^{2} \sqrt{2 \pi n}$ to obtain
$$
\frac{12n - 4}{(12n - 3) \sqrt{2 \pi} \cdot n^{\frac{5}{2}}}
\le \frac{n^{n - 2} e^{-n}}{n!}
\le \frac{12n + 6}{(12n + 7) \sqrt{2 \pi} \cdot n^{\frac{5}{2}}}
\le \frac{1}{\sqrt{2 \pi}} \cdot n^{-\frac{5}{2}}
$$
Now summing over all $n > 0$ and using the $p$-series and comparison test yields the convergence.

Using Stirlings approximation $n! \sim \sqrt{2 \pi n} \cdot n^n e^{-n}$ we have
$$
\sum_{n = 1}^{\infty} \frac{n^{n - 2} e^{-n}}{n!}
\sim \sum_{n = 1}^{\infty} \frac{n^{n - 2} e^{-n}}{\sqrt{2 \pi n} \cdot n^n e^{-n}}
= \sum_{n = 1}^{\infty} \frac{1}{\sqrt{2 \pi n} \cdot n^2}
= \sum_{n = 1}^{\infty} \frac{n^{-\frac{5}{2}}}{\sqrt{2 \pi}},
$$
which again converges because of the $p$-series test.
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