Hello there I want to find a proof for the generating formula for odd values of Dirichlet beta function given by wikipedia: link I searched MSE and didnt find something similar.

My try was to start with the cosine infinite product $$\cos{(\pi x)}=\prod_{n\ge 1} \left(1-\frac{4x^2}{(2n-1)^2}\right)\;,$$ take logarithms and differentiate it to get: $$-\pi\tan{(\pi x)}= \sum_{n\ge 1} \frac{\frac{-8x}{(2n-1)^2}}{1-\frac{4x^2}{(2n-1)^2}}$$ Then multiplying by x and expanding the denominator into a geometric series yields: $$\pi x\tan{(\pi x)}= 2\sum_{n\ge 1} \sum_{k\ge 0} \left(\frac{2x}{(2n-1)}\right)^{2k}\frac{4x^2}{(2n-1)}=2\sum_{n\ge 0} \sum_{k\ge 0} \left(\frac{2x}{(2n+1)}\right)^{2k+1}$$ Now already in this post: link a way has been shown to expand $\tan(x)$ into a power series, but I have $x\tan(x)$ and I don't know how to equate the 'coefficients' this way. Can I get some help with this? Also, what would be the relationship between Bernoulli numbers and Euler numbers in order to achieve the final result given by Wikipedia?

EDIT: Related: Euler numbers grow $2\left(\frac{2}{ \pi }\right)^{2 n+1}$-times slower than the factorial? However I am still interested in how to show this using my idea.

  • 4
    $\begingroup$ You made a mistake during your derivation, the correct one should be: $$\pi x\tan{(\pi x)}= 2\sum_{n\ge 1} \sum_{k\ge 0} \left(\frac{2x}{(2n-1)}\right)^{2k}\frac{4x^2}{(2n-1)^\color{red}{2}}$$ hence all you got is the usual $\zeta(2n+2)$, $\endgroup$ – pisco Jul 2 '18 at 7:08
  • $\begingroup$ Possible duplicate of Infinite Series $\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}$ $\endgroup$ – mrtaurho Jan 30 at 9:59
  • $\begingroup$ I would especially recommend to read robjohn's answer since it does not rely that heavily on complex analysis but answers the question quite well. $\endgroup$ – mrtaurho Jan 30 at 10:01
  • 1
    $\begingroup$ @Zacky I am sorry that I misread your post; I thought you would be only interested in a proof at all. $\endgroup$ – mrtaurho Jan 30 at 10:09
  • 1
    $\begingroup$ @Zacky I would suggest: leave it where it is. Even though your method does not turn out to work it is interesting enough seeing another try here on MSE. Furthermore I stumbled upon your post while actually searching for a proof of this formula. Hence your post is easily to be found I guess it might be of advantage leaving the post here. $\endgroup$ – mrtaurho Jan 30 at 10:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.