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 ?
 A: Note that
$$
\frac{n^{n-2}}{e^n n!}
=\frac{1}{n^2}\frac{n^n}{e^n n!}
=\frac{1}{n^2}a_n,
\quad \text{where }
a_n
=\frac{n^n}{e^n n!}.
$$
Note also
$$
\frac{a_{n+1}}{a_n}
=\cfrac{\frac{(n+1)^{n+1}}{e^{n+1} (n+1)!}}{\frac{n^n}{e^n n!}}
=\frac{1}{e}\left(1+\frac{1}{n}\right)^n
<1,
$$ which means that the sequence $(a_n)$ is decreasing, so $a_n<a_1$ and thus the series is dominated by the (convergent) series $\sum_{n=1}^{\infty}\frac{1}{n^2}a_1$ so it must be convergent, too.
A: Convergence
As I mentioned in a comment, Theorem $4$ from this answer shows that
$$
\frac{\,n^{n-2}}{e^nn!}\le\frac1{\sqrt{2\pi}\,n^{5/2}}\tag1
$$
Thus, by comparison with $\frac1{n^{5/2}}$, the series in question converges by the $p$-test ($p=5/2\gt1$). The $p$-test is proven as an example of the Cauchy-Condensation Test using a result about geometric series that can be proven using the Ratio Test, but is not given a name there.

Value
We can actually compute the value of the sum as follows.
Using the Taylor series for the Lambert W function derived in this answer, we see that
$$\newcommand{\W}{\operatorname{W}}
-\W(-x)=\sum_{k=1}^\infty\frac{n^{n-1}}{n!}x^n\tag2
$$
With $u=-\W(-x)$, we get $x=ue^{-u}$, and therefore,
$$
\begin{align}
\sum_{n=1}^\infty\frac{n^{n-2}}{e^nn!}
&=\int_0^{1/e}\frac{-\W(-x)}x\,\mathrm{d}x\tag3\\
&=\int_0^1(1-u)\,\mathrm{d}u\tag4\\[6pt]
&=\frac12\tag5
\end{align}
$$
A: 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.
