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The Catalan Zeta Function is defined as $$ \beta(n) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^n}. $$ For example, we have $\beta(1) = \pi/4$, $\beta(3)=\pi^3/32$, $\beta(5)=5 \pi^5/1536$.

I am trying to prove this identity $$ \beta(2n+1) = (-1)^n \left(\frac \pi 2 \right)^{2n+1}\frac{E_{2n}}{2 (2n)!}, $$ where $E_n$ denotes Euler numbers.

As I know that $$\sec z=1+\frac{z^{2}}{2}+\frac{5}{24}z^{4}+\frac{61}{720}z^{6}+\cdots+\frac{(-% 1)^{n}E_{2n}}{(2n)!}z^{2n}+\cdots.$$ I only need to show that $$ \sec\left(\frac{\pi x}{2}\right) = \sum_{n \ge 0} \frac{\beta(2n+1)}{\pi/4} x^{2n}. $$ But I got stuck here.

The question comes from The Computer as Crucible (see pp. 60). A clue is to use "an appropriate product for $cos(\pi x / 2)$".

NonalcoholicBeer
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  • Why not just use good ol' induction? – Simply Beautiful Art Jun 11 '17 at 19:01
  • $\beta(s) = \sum_{k=0}^\infty (-1)^k (2k+1)^{-s}$ is the Dirichlet beta function. There is no known closed-form for $s= 2n$ – reuns Jun 11 '17 at 19:03
  • @user1952009 Yes... and that relates... how? – Simply Beautiful Art Jun 11 '17 at 19:06
  • https://arxiv.org/pdf/math/0101168.pdf ... is worth a gander. – Donald Splutterwit Jun 11 '17 at 19:09
  • You can prove the last equation starting from the partial fraction decomposition of $\sec$, https://math.stackexchange.com/questions/210698/partial-fraction-of-secz-from-frac-pi-sin-pi-z –  Jun 11 '17 at 19:09
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    Catalan zeta, Dirichlet beta... gosh, I'd call it the $L$-function of the nontrivial character mod $4$. In any case, computing its explicit values at odd exponents should run pretty closely to calculations of the Riemann zeta-function at even integers. Sit down with a proof of the Riemann zeta evaluation and try to modify the steps to fit this Dirichlet series instead. – KCd Jun 11 '17 at 19:21
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    It's calculating $L(s,\chi)$ for odd positive integer $s$ where $\chi$ is the Dirichlet character of conductor $4$. One can do this by evaluating $L(s,\chi)$ at negative integers and using the functional equation. – Angina Seng Jun 11 '17 at 19:30
  • Have a look at this question: https://math.stackexchange.com/questions/850442/an-interesting-identity-involving-powers-of-pi-and-values-of-etas – Jack D'Aurizio Jun 11 '17 at 19:55

2 Answers2

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Show that $$f(z) =\frac{\pi^2\sin(\pi z)}{\cos^2(\pi z)}=\sum_{k=-\infty}^\infty \frac{(-1)^k}{(z+k+1/2)^2} $$ ($\frac{\pi^2\sin(\pi z)}{\cos^2(\pi z)}-\sum_{k=-\infty}^\infty \frac{(-1)^k}{(z+k+1/2)^2}$ is a bounded entire function vanishing at $i\infty$)

and look at $f^{(2n-1)}(0) = (2n+1)! 2^{2n+2} \sum_{k=0}^\infty \frac{(-1)^{k+1}}{(2k+1)^{2n+1}}$.

Also $f(z)$ is the derivative of $\frac{\pi}{\cos(\pi z)}$ (related to the Euler numbers)

reuns
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By the Mittag-Leffler expansion, $$ \sec{\left(\frac{\pi z}{2}\right)} = 1+\frac{2}{\pi}\sum_{n=-\infty}^{\infty} (-1)^{n-1}\left(\frac{1}{z-(2n+1)}+\frac{1}{2n+1}\right) = \frac{4}{\pi} \sum_{n=0}^{\infty} (-1)^{n-1} \frac{2n+1}{z^2-(2n+1)^2}. $$ You can then expand the fraction about $z=0$ to get $$ \frac{2n+1}{z^2-(2n+1)^2} = \frac{1}{2n+1} \frac{1}{1-\left( z/(2n+1) \right)^2} = \sum_{k=0}^{\infty} \frac{z^{2k}}{(2n+1)^{2k+1}} $$ and interchanging the order of summation will give the result.

Chappers
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