Why the step $\sum\limits_n n^{-s}=\sum\limits_n n(n^{-s}-(n+1)^{-s})$?

Question

I don't quite follows this step on a proof:

$$\sum_n n^{-s}=\sum_n n(n^{-s}-(n+1)^{-s})$$

I'm sure it's quite simple, but I'm afraid I don't see it.

Answer 1

Assume $\text{Re}(s)>1$. By using Abel's summation by parts, $$ \sum_{n=1}^N f_n(g_{n+1}-g_n) = \left(f_{N+1}g_{N+1} - f_1 g_1\right) - \sum_{n=1}^N g_{n+1}(f_{n+1}- f_n) $$ applied to $f_n=n$, $g_n=\dfrac1{n^s}$, one gets $$ \begin{align} \sum_{n=1}^N n\left(\frac1{(n+1)^s}-\frac1{n^s}\right) &= \left[(N+1)\frac1{(N+1)^s} - 1\right] - \sum_{n=1}^N \frac1{(n+1)^s} \\\\&=\frac1{(N+1)^{s-1}}-\sum_{n=1}^{N+1} \frac1{n^s} \end{align} $$ then, by letting $N \to \infty$, one obtains

$$ \sum_{n=1}^\infty n\left(\frac1{n^s}-\frac1{(n+1)^s}\right)=\sum_{n=1}^\infty \frac1{n^s}=\zeta(s),\qquad \text{Re}(s)>1, $$

as announced.