WHat is the inner steps here?

Question

I was studying abour Riemann zeta function over here where in (3) it has been written "by Abel's theorem", we have $$\sum\limits_{n\geq 1}\frac{1}{n^s}=\sum\limits_{n\geq 1}n\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right)$$ The remaining part I understood properly but this one I could not make out. Can you please help me to make it clear to me ?

Thank you

Answer 1

Verify the following:

$$\begin{array}{ll} \displaystyle \sum_{n=1}^N n(a_n-a_{n+1}) & =(a_1-a_2)+2(a_2-a_3)+\cdots+N(a_N-a_{N+1}) \\ & = (a_1+a_2+\cdots+a_N)-Na_{N+1}. \end{array}$$

Apply with $a_n:=n^{-s}$ and take $N\to\infty$.

Answer 2

Assume $\Re s>1$.

You can use Abel's theorem or you can proceed as follows, for $N\geq1$, $$ \begin{align} \sum_{n= 1}^{N}n\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right)&=\sum_{n= 1}^{N}\frac{1}{n^{s-1}}-\sum_{n= 1}^{N}\frac{n+1-1}{(n+1)^s}\\\\ &=\sum_{n= 1}^{N}\frac{1}{n^{s-1}}-\sum_{n= 1}^{N}\frac{1}{(n+1)^{s-1}}+\sum_{n= 1}^{N}\frac{1}{(n+1)^{s}}\\\\ &=\sum_{n= 1}^{N}\frac{1}{n^{s-1}}-\sum_{n= 2}^{N+1}\frac{1}{n^{s-1}}+\sum_{n= 2}^{N+1}\frac{1}{n^{s}}\\\\ &=1-\frac{1}{(N+1)^{s-1}}+\sum_{n= 2}^{N+1}\frac{1}{n^{s}}\\\\ &=-\frac{1}{(N+1)^{s-1}}+\sum_{n=1}^{N+1}\frac{1}{n^{s}} \end{align} $$ and let $N \to +\infty$ to get the announced result.