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 $\zeta(2)$
$\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} $