# What is $f(2s+1)$ when $f(s)=\sum_{n=0}^\infty {\frac{(-1)^n}{(2n+1)^s}}=1-\frac{1}{3^s}+\frac{1}{5^s}-\frac{1}{7^s}+\dots$? [duplicate]

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Is there an exact form of $$f(s)=\sum_{n=0}^\infty {\frac{(-1)^n}{(2n+1)^s}}=1-\frac{1}{3^s}+\frac{1}{5^s}-\frac{1}{7^s}+\dots$$ when $s$ is odd?

Discussion

I have been exploring infinite series and will be spending my evening looking for patterns in this particular class. I invite the interested reader to join me and the not so interested to just move along. I will be updating this question with relevant facts as the evening unfolds.

There must (there must!) be some closed form in terms of $\pi$ and $s$ when $s$ is odd and there will definitely be something we can say about how this relates to the generalized zeta function.

$f(1)=\frac{\pi}{4}$

$f(2)=$Catalan. [I will leave a remark about this below.]

$f(3)= \frac{1}{64} (ζ(3, 1/4) - ζ(3, 3/4))=\frac{\pi^3}{32}$

$f(4)= \frac{1}{256} (ζ(4, 1/4) - ζ(4, 3/4))$

$f(5)= \frac{1}{1024}(ζ(5, 1/4) - ζ(5, 3/4)) =\frac{5\pi^5}{1536}$

$f(6)= \frac{1}{4096}(ζ(6, 1/4) - ζ(6, 3/4)$

$f(7)=\frac{1}{16384}(ζ(7, 1/4) - ζ(7, 3/4))= \frac{61 π^7}{184320}$

I thought about posting in Meta asking about this type of question. It's a "call to adventure" question: Come look at this with me if you so please. If you're not into it... downvote the question/let me know in the comments/move on to some other question that you do enjoy.

Update 1: It looks like $$f(s)= \frac{1}{2^{2s}} \Bigg(\zeta(s, \frac{1}{4})-\zeta(s, \frac{3}{4}) \Bigg)$$

Update 2: A remark on Catalan's number and on $s$ even in general. The wiki page claims it to be unknown whether this Catalan's constant is irrational or transcendental. Come on guys? What do we pay you for? Let me just state for the conjectural record that $\sum_{n=1}^\infty\frac{a_n}{n^s}$ for a periodic sequence of integers $a_n$ has just must be transcendental (it must!). I am very confident this is the case when $a_n$ has period of prime $p$ and for $s=1$. It's surprising to me that I would need these conditions. Note that for $f(2)$ the numerators of the series would be $1,0,-1,0 \dots$ and that's not a prime period and also $s \neq 1$ so we cannot use any of those tools to make any statements about Catalan's number but also... one cannot deny the conjecture isn't really too bold. Most numbers should be transcendental and this periodic numerators of these series must be a push in the transcendental direction.

## marked as duplicate by Simply Beautiful Art, Community♦Aug 17 '18 at 1:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• chat.stackexchange.com/rooms/81142/mason-korb – Mason Aug 4 '18 at 2:38
• Does the sum start at $n=0$ or $n=1$? – Simply Beautiful Art Aug 4 '18 at 2:46
• Should be $n=0$. But this said $n=1$ originally. I edited. But apparently not the title. Thanks. I changed it. – Mason Aug 4 '18 at 2:47
• The result follows from the functional equation of $L(s, \chi_4)$ where $\chi_4(n) = 1$ if $n\equiv 1 (4)$, $-1$ if $n\equiv -1(4)$. – Sungjin Kim Aug 4 '18 at 3:01
• Note this is the Dirichlet beta function. – Simply Beautiful Art Aug 4 '18 at 4:04

## 2 Answers

As you've noticed, we have

$$4^sf(s)=\sum_{n=0}^\infty\left(\frac1{(n+1/4)^s}-\frac1{(n+3/4)^s}\right)$$

which, for odd $s$, can be written as

\begin{align} &(s-1)!4^sf(s)\\ &=\lim_{x\to1/4}\frac{d^{s-1}}{dx^{s-1}}\sum_{n=0}^\infty\left(\frac1{n+x}+\frac1{x+1-n}\right)\\ &=\lim_{x\to1/4}\frac{d^{s-1}}{dx^{s-1}}\sum_{n=-\infty}^\infty\frac1{n+x} \\ &=\lim_{x\to1/4}\frac{d^{s-1}}{dx^{s-1}}\pi\cot(\pi x) \end{align}

which gives exact values for $f(s)$ when $s$ is odd.

It also turns out this is the Dirichlet beta function with the special values of

$$f(s)=\beta(s)=\frac{(-1)^kE_{2k}}{4^{k+1}(2k)!}\pi^{2k+1}$$

where $s=2k+1$ and $E_k$ are the Euler numbers.

• What happens when $s$ is even? Are those values not exact also? – Mark Viola Aug 4 '18 at 3:10
• The operations $\lim_{x\rightarrow 1/4}$, $d^{s-1}/dx^{s-1}$ and $\sum_{n=0}^{\infty}$ have been interchanged. These interchanges must be done carefully at each step. – Sungjin Kim Aug 4 '18 at 3:15
• Why does your development fail for even values of $s$? – Mark Viola Aug 4 '18 at 3:26
• @MarkViola one can write it as $$\frac{4^sf(s)}{(s-1)!}=\psi^{(s-1)}(1/4)-\psi^{(s-1)}(3/4)$$using polygamma functions, which can be reduced to evaluating either polygamma function due to the reflection formula (or by following the process in my answer), reducing it to finding $\sum_{n=0}^\infty\frac1{(4n+1)^s}$, which doesn't reduce any further AFAIK. – Simply Beautiful Art Aug 4 '18 at 3:32
• As per why we cannot just apply the result from my answer, note the derivatives of $\frac1{n+x}$ alternates signs, while the derivatives of $\frac1{n-x}$ do not. So the signs only line up nicely for odd $s$. – Simply Beautiful Art Aug 4 '18 at 3:34

You can write this as $$-\frac{i}2 \left( {\it polylog} \left( s,i \right) -{\it polylog} \left( s,- i \right) \right)$$

EDIT: It does seem that for odd $s$, $f(s)$ is a rational multiple of $\pi^s$. See OEIS sequence A053005 and A046976 and references there.

• The answer has been improved a bit, but there is still no explanation, and the only parts of the answer that provides some explanation are links. I shouldn't have to mention the help center, not to a user such as you. Until then, I'll keep my downvote. – Simply Beautiful Art Aug 4 '18 at 4:18
• en.wikipedia.org/wiki/Polylogarithm – Mason Aug 4 '18 at 23:56