another funky identity

Start with this:

\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...

 

Advertisements