need help understanding a solution of sequence and series

Question

On a particular problem of integration a line came up like this-- $$S_{n}=\frac{k}{k-1}\sum_{m=1}^{\infty}\left(\frac{1}{m^{k-1}}-\frac{1}{(m+1)^{k-1}}\right)+\sum_{m=1}^{\infty}m \left[\frac{1}{(m+1)^k}-\frac{1}{m^k}\right]$$ From this point they directly wrote the summation as $$S_{n}=\frac{k}{k-1}-\zeta(k)$$ I don't understand how they got this identity

Answer 1

If $k > 1$, this series is telescoping $$\sum_{m=1}^{\infty}\left(\frac{1}{m^{k-1}}-\frac{1}{(m+1)^{k-1}}\right) = 1$$ and $$\sum_{m=1}^{\infty}m\left(\frac{1}{m^{k}}-\frac{1}{(m+1)^{k}}\right) = \sum_{m=1}^{\infty}\left(\frac{m}{m^{k}}-\frac{m+1}{(m+1)^{k}}+\frac{1}{(m+1)^{k}}\right) = 1+(\zeta(k)-1)$$

Answer 2

Note that

$$\begin{align} \sum_{m=1}^\infty m\left(\frac{1}{(m+1)^k}-\frac{1}{m^k}\right)&=\sum_{m=1}^\infty \frac{m}{(m+1)^k}-\sum_{m=1}^\infty\frac{m}{m^k}\\\\ &=\sum_{m=1}^\infty \frac{m-1}{m^k}-\sum_{m=1}^\infty\frac{m}{m^k}\\\\ &=-\sum_{m=1}^\infty\frac{1}{m^k}\\\\ &=-\zeta(k) \end{align}$$