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I've found this identity of a blog: $$\sum _{n=1}^{\infty } \frac{\left(\frac{4}{9}\right)^n \beta (2 n+1)}{n+1}=\frac{-9 \left(-\frac{2 C}{3}-1+\frac{1}{3} \pi \log \left(\sqrt{3}+2\right)\right)}{2 \pi }$$ Where $\beta$ denotes Dirichlet Beta function (i.e. A L-function modulo 4). Unfortunately, this result seems to be incorrect: numerical approximations failed disastrously. My question is: Is there a closed-form for this sum? After all, there are so many Zeta sums with fabulous closed-forms (see e.g. Srivastava's review), and this made me wonder if the given result can be modified to be correct. I tried reverse the order of summation but it lead to nothing valuable. Therefore I'd like you to help me. Thank you all!

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Here is a sketch. Using, say, the link with Euler numbers, one gets $$\sum_{n=0}^{\infty}\beta(2n+1)x^{2n}=\frac{\pi}{4}\sec\frac{x\pi}{2},$$ which implies $$\sum_{n=0}^{\infty}\frac{\beta(2n+1)}{2n+2}x^{2n+2}=\frac{1}{\pi}f\left(\frac{x\pi}{2}\right),\quad f(y)=\int_0^y\frac{x\,dx}{\cos x}.$$ The given sum is then equal to $\dfrac{9}{2\pi}f\left(\dfrac{\pi}{3}\right)-\dfrac{\pi}{4}$, with the known value of $$f\left(\frac{\pi}{3}\right)=\frac{\pi}{3}\ln(2+\sqrt{3})-\frac{2}{3}G,$$ where $G$ is Catalan's constant. The latter formula can be derived from $$\int_0^y\ln\left|\tan\frac{x}{2}\right|\,dx=-2\sum_{n=0}^{\infty}\frac{\sin(2n+1)y}{(2n+1)^2}.$$

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  • $\begingroup$ Is your $G$ the same as OP's $C$? $\endgroup$ – Gerry Myerson Oct 17 '19 at 11:26
  • $\begingroup$ A neat but clever answer. Thank you! $\endgroup$ – 雄凤山 Oct 17 '19 at 15:03
  • $\begingroup$ @GerryMyerson Yes it is. $\endgroup$ – 雄凤山 Oct 17 '19 at 15:04

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