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

Question

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.

Answer 1

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.

Answer 2

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.