another funky identity

$\sum_{n=1}^{\infty} \frac{e^{1/n}}{n^{2}}$

Which can be rewritten:

$\sum_{n=1}^{\infty} \frac{1}{n^{2}} \sum_{m=0}^{\infty} \frac{(1/n)^{m}}{m!}$

Rearranging:

$\sum_{m=0}^{\infty} \frac{1}{m!} \sum_{n=1}^{\infty} \frac{1}{n^{m+2}}$

The inside is simply a zeta function, so finally we have:

$\sum_{m=0}^{\infty} \frac{\zeta(2+m)}{m!}$

Therefore:

$\sum_{n=1}^{\infty} \frac{e^{1/n}}{n^{2}} = \sum_{m=0}^{\infty} \frac{\zeta(2+m)}{m!}$ according to Wolfram Alpha, this is about

3.6133873336591394587596763...