TITLE: Prove that this summation evaluates out to $\zeta(2)-1$
QUESTION [5 upvotes]: I am aware of the following identity:
$$\sum_{m=1}^\infty \left(\frac{1}{m}-\left(\zeta(2)-\sum_{n=1}^m \frac{1}{n^2}\right)\right)=\zeta(2)-1$$
I can't quite figure out how to prove this result. Maybe this has to do with some specific properties of $\zeta(x)$, but I really don't know enough yet. If possible, as well, could one generalize this result for looking at more than just $\zeta(2)$

REPLY [5 votes]: $\sum_{m=1}^\infty \left(\frac{1}{m}-\left(\zeta(2)-\sum_{n=1}^m \frac{1}{n^2}\right)\right)=\zeta(2)-1
$
Playing around
and seeing what happens.
Since
$\zeta(2)
=\sum_{n=1}^{\infty} \frac{1}{n^2}
$,
the left side is
$\begin{array}\\
\sum_{m=1}^\infty \left(\frac{1}{m}-\left(\sum_{n=m+1}^{\infty} \frac{1}{n^2}\right)\right)
&=\sum_{m=1}^\infty \left(\sum_{n=m+1}^{\infty}\left(\frac{1}{n-1}-\frac1{n}\right)-\left(\sum_{n=m+1}^{\infty} \frac{1}{n^2}\right)\right)\\
&=\sum_{m=1}^\infty \left(\sum_{n=m+1}^{\infty}\left(\frac{1}{n(n-1)}\right)-\left(\sum_{n=m+1}^{\infty} \frac{1}{n^2}\right)\right)\\
&=\sum_{m=1}^\infty \left(\sum_{n=m+1}^{\infty} \left(\frac{1}{n(n-1)}-\frac{1}{n^2}\right)\right)\\
&=\sum_{m=1}^\infty \sum_{n=m+1}^{\infty} \frac{1}{n^2(n-1)}\\
&=\sum_{n=2}^{\infty}\sum_{m=1}^{n-1}  \frac{1}{n^2(n-1)}
\qquad\text{(Looking good!)}\\
&=\sum_{n=2}^{\infty}(n-1)  \frac{1}{n^2(n-1)}\\
&=\sum_{n=2}^{\infty} \frac{1}{n^2}\\
&=\zeta(2)-1
\qquad\text{Shazam!}
\end{array}
$