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

Find the value of the closed form of the above series to 3 decimal places.

The answer is 1.645.

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The answer is $\frac{\pi^{2}}{6}$ .

$\sum_{n=1}^\infty \frac{n}{m^{n+1}}$ can be written as $\frac{\sum_{n=0}^\infty \frac{n}{m^{n}}}{m}$ , or $\frac{\sum_{n=0}^\infty \frac{n+1}{m^{n+1}}}{m}$ , as the first term is $0$ .

It can further be written as $\frac{\sum_{n=0}^\infty \frac{n}{m^{n}}}{m^{2}} + \frac{\sum_{n=0}^\infty \frac{1}{m^{n}}}{m^{2}}$ .

Simplifying, we get $\frac{\sum_{n=0}^\infty \frac{n}{m^{n}}}{m} = \frac{1}{(m-1)^{2}}$ , whose summation from 2 to $\infty$ is $\frac{\pi^{2}}{6}$ .