Step in proof of Dirichlet Series Convergence

Question

I am reading this proof, and am wondering about the justification of the step labeled "telescoping the series on the right."

$$\sum_{n=m}^{N-1} |(n+1)^{-s}-n^{-s}| \le |N^{-s}-m^{-s}|$$ where $\operatorname{Re}(s)>0$.

Why are you allowed to do this when there are absolute value signs?

I don't know where in the proof you linked is the expression above, but in what you wrote everything's positive for any $;s\in\Bbb R;$ ... — DonAntonio, Nov 18 '13 at 16:01

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

Good point, which I think the author of ProofWiki proof has missed. When $\operatorname{Re} s>0$, we have

$$\begin{split} \left|\frac{1}{(n+1)^{s}}-\frac{1}{n^s}\right| & =\left|\int_{n}^{n+1}\frac{s}{x^{s+1}}\,dx\right| \\ & \leq |s| \int_{n}^{n+1}\frac{dx}{x^{\operatorname{Re} s +1}} \\ &=\frac{|s|}{\operatorname{Re} s }\left(\frac{1}{n^{\operatorname{Re} s }}-\frac{1}{(n+1)^{\operatorname{Re} s }}\right) \end{split}$$

Which leads to telescoping, but with the extra constant factor in front.

As seen in Uniform convergence of Dirichlet series, answer by Josué Tonelli-Cueto.

Step in proof of Dirichlet Series Convergence

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