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Let $s \in \mathbb C$ be such that the series $$ \sum\limits_{n=1}^{\infty} \frac {(-1)^n} {n^{\mathfrak {R} (s)}} $$ converges. From here can it be concluded that $$ \sum\limits_{n=1}^{\infty} \frac {(-1)^n} {n^s} $$ also converges?

Is the same true for divergence also? Actually I need this to conclude that the series $$ \sum\limits_{n=1}^{\infty} \frac {(-1)^n} {n^s} $$ converges iff $\mathfrak {R} (s) \gt 0$. Could anyone please help me?

Thanks a bunch!

Daniele Tampieri
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Fanatics
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  • Have you tried collecting pairs of alternating terms and seeing what happens with the resulting series? – Anonymous Aug 22 '21 at 06:36
  • $-\eta (s)=-\left(1-2^{1-s}\right) \zeta (s)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \frac{(-1)^n}{n^s}\right),\Re(s)>0$ (see https://en.wikipedia.org/wiki/Dirichlet_eta_function). – Steven Clark Aug 22 '21 at 15:38
  • @Steven Clark I can't find any proof of the fact that $\eta (s)$ converges for $\mathfrak {R} (s) \gt 0$ in the wikipedia link that you have provided. – Fanatics Aug 22 '21 at 16:04
  • If you search here for "proof of convergence of Dirichlet eta function Dirichlet series" you'll find your question has been asked and answered before (e.g. https://math.stackexchange.com/questions/1042512/proof-of-convergence-of-dirichlets-eta-function). – Steven Clark Aug 22 '21 at 16:13
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    @StevenClark Do you understand why the question you linked to refers to the "alternating zeta function" but then doesn't include a $(-1)^n$ in the term? And do you consider the answer there to be as simple as it should be? – Anonymous Aug 22 '21 at 17:16
  • @Anonymous I believe the missing $(-1)^{n+1}$ is a typo. The Dirichlet series $\zeta(s)=\sum\limits_{n=1}^\infty\frac{1}{n^s}$ only converges for $\Re(s)>1$. I don't think the answer I pointed to is the simplest, but I just pointed to the first answer I found and figured the OP could search Math StackExchange and more generally the Internet for Dirichlet eta function to find various proofs. – Steven Clark Aug 22 '21 at 17:51

1 Answers1

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Collect alternating pairs of terms. The sum of such a pair is

$$(2k)^{-s} - (2k - 1)^{-s} = (2k)^{-s}[1 - (1 - 1/2k)^{-s}] \sim -s(2k)^{-s-1}. $$

The series with this general term converges absolutely whenever $\Re(s) > 0$. Since moreover the general term of the original series tends to zero, this proves convergence when $\Re(s) > 0$.

Daniele Tampieri
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Anonymous
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  • The series $\sum\limits_{k=1}^{\infty} \frac {1} {(2k)^{s+1}}$ converges absolutely if $\mathfrak {R} (s)\gt 0$ and hence converges in the right half plane excluding the imaginary axis. Isn't it so? – Fanatics Aug 22 '21 at 17:14
  • Right. I didn't discuss the case $R(s) \leq 0$. You need to do that separately. – Anonymous Aug 22 '21 at 17:17
  • If $\mathfrak {R} (s) \leq 0$ the same reasoning wouldn't apply. Because a series can be convergent without being absolutely convergent which are called conditionally convergent series. – Fanatics Aug 22 '21 at 17:22
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    Right, but in that case there's a simple argument to show the original Dirichlet series doesn't converge. – Anonymous Aug 22 '21 at 17:25
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    Look, this part of the question is much easier than the rest. If you think about it, I'm sure you'll see. – Anonymous Aug 22 '21 at 17:30
  • Will it be like the limit of the general term doesn't converge to $0\ $? – Fanatics Aug 22 '21 at 17:30
  • Right, the term doesn't converge to zero. – Anonymous Aug 22 '21 at 17:31
  • When you are saying that the $k$-th term of the given series is approximately equal to the $k$-th the term of the series $s \sum\limits_{n=1}^{\infty} \left ( \frac {1} {2n} \right )^{s+1}$ does it allow you to characterize the convergence of the given series in terms of approximated series? For that you need the difference of their tail to be less than given $\varepsilon.$ Can it be guaranteed anyway? – Anil Bagchi. Aug 24 '21 at 15:26
  • @AntonioClaire What I mean by the symbolism $a_k \sim b_k$ is that $a_k/b_k \to 1$ as $k \to +\infty$. (In this case, this can be proved by using a first-order Taylor expansion of $(1+x)^{-s}$.) When this occurs, the absolute convergence of one of the series $\sum a_k$, $\sum b_k$ implies absolute convergence of the other. – Anonymous Aug 24 '21 at 16:06
  • Do you use limit form of the comparison test? – Anil Bagchi. Aug 24 '21 at 16:09
  • Actually what you need is the convergence; not the absolute convergence. – Anil Bagchi. Aug 24 '21 at 16:15
  • If you assume that $\sum a_k$ converges absolutely, then under the hypothesis $a_k \sim b_k$, you have $|b_k| \leq C|a_k|$ for some constant $C$ and sufficiently large $k$, so $\sum |b_k|$ converges too. This is the comparison test used. Absolute convergence is a necessary hypothesis. For example, $(-1)^k/\sqrt{k} \sim (-1)^k/\sqrt{k} + 1/k$. The series with the first term converges, but not the second. – Anonymous Aug 24 '21 at 16:26
  • @Anonymous that's exactly my point. The OP doesn't want abscissa of absolute convergence. Rather he wants abscissa of convergence. So I don't think that this answer suits his purpose. – Anil Bagchi. Aug 24 '21 at 16:35
  • @Antonio Claire I think you're misunderstanding my argument. Letting $c_n = (-1)^n/n^s$, what I prove is that the series $(c_1 + c_2) + (c_3 + c_4) + (c_5 + c_6) + \dots$ converges absolutely., not the series $c_1 + c_2 + c_3 + \dots$. I am deducing from the previous fact that this series converges, but I am not saying it converges absolutely. This is what I meant by "Collect alternating pairs of terms." – Anonymous Aug 24 '21 at 17:00
  • Sorry I am busy with some other works. That's why I am responding late. I understand that you proved that $(c_1 + c_2) + (c_3 + c_4) + (c_5 + c_6) + \cdots$ converges absolutely. But from here how does it follow that $\sum c_i$ converges? The easiest counter-example is when $c_n = (-1)^n,$ for all $n \in \mathbb N.$ If you collect alternating pairs then they are all zero and hence the sum of alternating pairs is absolutely convergent but $\sum c_i$ is clearly not convergent. – Anil Bagchi. Aug 26 '21 at 06:23
  • @AntonioClaire Let $S_{n}$ be the sequence of partial sums of $\sum c_n$. Then the convergence of the series formed from the collected pairs amounts to the convergence of the sequence $S_{2n}$. But since $S_{2n+1} - S_{2n} = c_{2n+1} \to 0$ (in view of $R(s)>0$), the sequence $S_{2n+1}$ converges to the same limit as $S_{2n}$. I explicitly point this out in my answer when I say, "Since moreover the general term of the original series tends to zero, this proves convergence when $R(s) > 0$." – Anonymous Aug 26 '21 at 06:39
  • @Anonymous now it's clear to me. This is certainly one of the most elementary answers to this question. Thanks a bunch! – Anil Bagchi. Aug 26 '21 at 07:39