Raabe-Duhamel test for $\sum_{n=1}^{+\infty}\frac{a^nn!}{n^n}$ I need to prove the convergence/divergence of this summation using Raabe-Duhamel test.
I reached to this point after writing up $\frac{a_n}{a_{n+1}}$:
$$\frac{(n+1)^n(n+1)}{an^n}$$
I don't know how to continue it.
$a\ge0$.

The ratio $a_n/a_{n+1}$ is the OP is not quite correct.  Instead, note that we have for $a\ne0$
$$\begin{align}
\frac{a_n}{a_{n+1}}&=\frac{a^nn!/n^n}{a^{n+1}(n+1)!/(n+1)^{n+1}}\\\\
&=\frac1a\,\,\frac{n!}{(n+1)!}\,\,\frac{(n+1)^{n+1}}{n^n}\\\\
&=\frac1a\,\,\frac1{n+1}\,\,\frac{(n+1)(n+1)^n}{n^n}\\\\
&=\frac1a\,\,\frac{(n+1)^n}{n^n}\\\\
&=\frac1a \left(1+\frac1n\right)^n
\end{align}$$
And inasmuch as $\lim\limits_{n\to \infty}\left(1+\frac1n\right)^n=e$, we find that
$$\lim_{n\to \infty }\frac{a_n}{a_{n+1}}=\frac ea$$
and the ratio test reveals that the series converges for $a<e$ and diverges for $a>e$.
But the ratio test is inconclusive for $a=e$.  For $a=e$ the Raabe-Duhamel test reveals
$$\lim_{n\to\infty}\left(n\left(\frac1e \left(1+\frac1n\right)^n-1\right) \right)=-\frac12$$
and we can conclude that the series diverges for $a\ge e$ and converges for $a<e$.
