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I would like to find a proof for the generating formula for odd values of Dirichlet beta function, namely: $$\beta(2k+1)=\frac{(-1)^kE_{2k}\pi^{2k+1}}{4^{k+1}(2k)!}$$

My try was to start with the cosine infinite product $$\cos{(\pi x)}=\prod_{n\ge 1} \left(1-\frac{4x^2}{(2n-1)^2}\right)\;,$$ Logarithmate and differentiate it, in order 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 it has been shown how to expand $\tan(x)$ into 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?

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.

Zacky
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    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 $\zeta(2n+2)$, – pisco Jul 02 '18 at 07:08
  • Possible duplicate of Infinite Series $\sum\limits_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}$ – mrtaurho Jan 30 '19 at 09:59
  • 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. – mrtaurho Jan 30 '19 at 10:01
  • It is not quite a duplicate, read the last line please: I am interested to solve this using my idea. However I think I should delete it, as said in the comments my idea leads to zeta function instead. – Zacky Jan 30 '19 at 10:04
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    @Zacky I am sorry that I misread your post; I thought you would be only interested in a proof at all. – mrtaurho Jan 30 '19 at 10:09
  • Well, no worries! I think the right answer would be that is impossible to do it my way (basically pisco's comment) among with my mistake which lead me to believe that there is the possibility. But I am thinking if this question is of any use at all.. – Zacky Jan 30 '19 at 10:16
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    @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. – mrtaurho Jan 30 '19 at 10:30

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