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in Titsmarsch book on zeta function page 17 he said that this series $\sum_{m=1}^\infty (-1)^m \{(m+1)^s - m^s\}$ is uniformly convergent on $0 < \Re(s) < 1$

for real $s$ i can see how it is but for complex $s$ i'm not sure can someone please help? thanks

john takeuchi
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1 Answers1

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Let $s\in \mathbb{C}$ and let $f(s)$ be given by

$$f(s)=\sum_{m=1}^\infty (-1)^{m}\left((m+1)^s-m^s \right) \tag 1$$

In THIS ANSWER, I used a Generalized Dirichlet Test (See here) to show that the alternating series representation $\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$ for the Dirichlet Eta function converges for $\text{Re}(s)>0$.

Noting that $\left((m+1)^s-m^s \right)=\frac s{m^{1-s}} +O\left(\frac1{m^{2-s}}\right)$, it is easily seen that the series on the right-hand side of $(1)$ converges for $0<\text{Re}(s)<1$.

Hence, by pairing successive even and odd terms, we can write

$$f(s)=\sum_{m=1}^\infty \left((2m+1)^s-2(2m)^s+(2m-1)^s\right)$$

Now, note that $\left((2m+1)^s-2(2m)^s+(2m-1)^s\right)=\frac{2^ss(s-1)}{4m^{2-s}}+O\left(\frac1{m^{4-s}}\right)$. Furthermore, for any $\delta>0$ and $0<\text{Re}(s)<1-\delta$ we have the estimates

$$\begin{align} \left|\sum_{m=1}^\infty \frac1{m^{2-s}}\right|&\le\sum_{m=1}^\infty \frac1{\left|m^{2-s}\right|}\\\\ &= \sum_{m=1}^\infty \frac{1}{m^{2-\text{Re}(s)}}\\\\ &\le \sum_{m=1}^\infty \frac{1}{m^{1+\delta}} \end{align}$$

This shows that the series in $(1)$ converges uniformly on all compact subsets contained on $0<\text{Re}(s)<1$.

Mark Viola
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  • thank you for the answer i have a few questions

    1. how did you get the big O especially for the $O \left( \frac{1}{m^{4-s}}\right)$

    2. is it ok to rearrange the terms like that since this is an infinite series?

    3. why is it only on compact subsets?

     – john takeuchi Apr 20 '20 at 21:26
  • You're welcome. My pleasure. The expansion comes from Taylor's Formula. The rearrangement was allowed since the series converges, – Mark Viola Apr 20 '20 at 21:30
  • Taylor expand $(m+1)^s-m^s$ w.r.t $m$ around 0? – john takeuchi Apr 20 '20 at 22:22
  • No, expand $$(2m+1)^s-(2m)^s+(2m-1)^s=(2m)^s\left(\left(1+\frac1{2m}\right)^s-2+\left(1-\frac1{2m}\right)^s\right)$$around $\frac1{2m}= 0$. – Mark Viola Apr 20 '20 at 22:54
  • thanks for the reply but im even more confused now $\frac{1}{2m}=0$? – john takeuchi Apr 20 '20 at 22:59
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    As $m\to \infty$, $\frac1{2m}\to 0$. Let $x=\frac1{2m}$ and write $$\begin{align}(1+x)^s-2+(1-x)^s&=1+sx+\frac12 s(s-1)x^2+\frac1{6}s(s-1)(s-2)x^3+O(x^4)\\& - 2\\&+1-sx+\frac12 s(s-1)x^2-\frac1{6}s(s-1)(s-2)x^3+O(x^4)\\&=s(s-1)x^2+O(x^4)\end{align}$$ – Mark Viola Apr 20 '20 at 23:05
  • ok i see it now let $\frac{1}{2m}=x$ then we have $\left ( \left (1+x\right)^s -2 +\left (1-x \right)^s \right)$ expand around $x=0$ and we evaluate at $x'=\frac{1}{2m}$ thanks again – john takeuchi Apr 20 '20 at 23:16
  • You're welcome. My pleasure. – Mark Viola Apr 20 '20 at 23:44
  • already did upvoted your answer and your comment :) thanks again – john takeuchi Apr 21 '20 at 00:15
  • your answer is already very good i actually used to do that approximation $\sqrt{1+\Delta} \sim 1 + \frac{1}{2}\Delta$ but never though of applying it to a series so that was a very good insight and thank you for that the only question i have is why does it apply only on compact subsets? – john takeuchi Apr 23 '20 at 01:20
  • For any $\delta >0$ the series uniformly converges for $0\le \text{Re}(s)\le 1-\delta$. – Mark Viola Apr 23 '20 at 02:13