I was computing an integral $I$ with a parameter $p$ and I found that $$ I\left(p\right)=\frac{p!}{\left(p-1\right)^{p+1}} $$
I was wondering if the series $\displaystyle \sum_{p \geq 2}^{ }I_p$ converges, i've found if I did not make mistake that $$ \frac{I\left(p+1\right)}{I\left(p\right)}\underset{p \rightarrow +\infty}{\rightarrow}\frac{1}{e} $$ Hence the sum should be convergent, how can I compute $\displaystyle \sum_{n=2}^{+\infty}I\left(p\right)$ ?
If you want we can write it as an integral. \begin{align} \sum_{p=2}^{+\infty}\frac{p!}{(p-1)^{p+1}}&=\sum_{p=1}^{+\infty}\frac{(p+1)!}{p^{p+2}}\\ &=\sum_{p=1}^{+\infty}\frac1p\int_0^{\infty}e^{-t}\left(\frac t p\right)^{p+1}dt\\ &=\sum_{p=1}^{+\infty}\int_0^{\infty}e^{-sp}s^{p+1}ds\tag{$t=sp$}\\ &=\int_0^{\infty}s\sum_{p=1}^{+\infty}(e^{-s}s)^{p}ds\\ &=\int_0^{\infty}\frac{s^2}{e^s-s}ds \end{align}