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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 ?

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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.

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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, $$ \begin{align} \sum_{n=1}^\infty\frac{n^{n-2}}{e^nn!} &=\int_0^{1/e}\frac{-\W(-x)}x\,\mathrm{d}x\tag3\\ &=\int_0^1(1-u)\,\mathrm{d}u\tag4\\[6pt] &=\frac12\tag5 \end{align} $$

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    $\begingroup$ Nicely done. One approach which I've tried (and failed) to make work is to replace $1/n!$ by an integral representation; the thought being to exchange the order of summation/integration, sum the resulting series, and then integrate. Alas, that doesn't seem to work out (I guess because the series is so close to being divergent?). $\endgroup$ Sep 6 '19 at 17:53
  • $\begingroup$ @Semiclassical: I had played with the series for Lambert W and I remembered trying to get the Taylor Series. I had gotten the recursion $(5)$, but had not yet applied $(7)$ from this answer to prove $(6)$. This answer completes the derivation of the series for Lambert W. $\endgroup$
    – robjohn
    Sep 6 '19 at 18:21
  • $\begingroup$ @ViktorGlombik: you had asked in a (deleted) comment to this question if I could prove that the sum of the series was $\frac12$, but I was not aware of your new question, so I posted the answer here. I will think about your new question. $\endgroup$
    – robjohn
    Sep 6 '19 at 18:29
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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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