We was able to determine $(1)$ to have this closed form

$$\ln(2)-\gamma=\sum_{n=1}^{\infty}{\zeta(2n+1)\over (2n+1)2^{2n}}\tag1$$

then we when on and try to evaluate $(2)$ and we only half of the closed form

$$2\ln(2)-\gamma-2X=\sum_{n=1}^{\infty}{\zeta(2n+1)\over (2n+1)2^{4n}}\tag2$$

Where $$X=\sum_{n=0}^{\infty}{\eta(2n+1)\over(2n+1)2^{2n+1}}\tag3$$

where $\eta$ is the Dirichlet eta function and $\gamma$ is Euler-Masheroni constant

How do we evaluate the closed form of $(3)?$

  • 2
    $\begingroup$ I am not sure, but have you tried the identity $$ \sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{2n+1} z^{2n+1} = -\gamma z + \frac{\log\Gamma(1-z) - \log\Gamma(1+z)}{2}$$? $\endgroup$ – Sangchul Lee Oct 7 '17 at 17:46
  • $\begingroup$ No I haven't @Sangchu Lee $\endgroup$ – gymbvghjkgkjkhgfkl Oct 7 '17 at 17:51
  • $\begingroup$ Hint: The generating funtion of Riemann-Zeta is given by $\gamma+\psi(1+x)=-\sum_{n\geq1}\zeta(n+1)(-x)^n$ $\endgroup$ – tired Oct 7 '17 at 18:40
  • $\begingroup$ going all the way through the algebra we obtain $-\gamma+2\log\left(\frac{\Gamma(3/4)}{\Gamma(5/4)}\right)$ i think $\endgroup$ – tired Oct 7 '17 at 19:02
  • $\begingroup$ Thank @tired. I was checking on the sum calculator the numerical value seem correct. $\endgroup$ – gymbvghjkgkjkhgfkl Oct 7 '17 at 19:12

We have the following Lemma (a sketch of a proof can be found below)

$$ s(x)=\sum_{n\geq 1}(-x)^n \zeta(n+1)=-\gamma-\psi(1+x)\quad \color{red}{(I)} $$

where $\psi(z)=\frac{d\log(\Gamma(z))}{dz}$ is the digamma function and $\gamma$ is Euler's constant

Integrating yields

$$ S(x)=\int dx s(x)=\sum_{n\geq1}\frac{\zeta(n+1)}{n+1}(-x)^{n+1}=-\gamma x-\log(\Gamma(1+x)) $$

Taking the odd part $$ S(x)-S(-x)=2\sum_{n\geq1}\frac{\zeta(2n+1)}{2n+1}x^{2n+1}=-2\gamma x-\log(\Gamma(1+x))+\log(\Gamma(1-x)) $$

Now let us put $x=\frac{1}{4}$ we get

$$ \sum_{n\geq1}\frac{\zeta(2n+1)}{2n+1}\frac1{4^{2n}}=-\gamma+2\log\left(\frac{\Gamma(3/4)}{\Gamma(5/4)}\right) $$

which is the sum of OP's interest

We now proof $\color{red}{(I)}$:

Use the definition of the $\zeta$-function as a series and exchange the order of summation. Doing the first sum yields $S(x)=\sum_{k\geq1}\frac{1}{x+k}-\frac{1}k$ expressing this in terms of Digamma functions yields $\color{red}{(I)}$.


  • $\begingroup$ I believe you meant to let $x=1/2$? $\endgroup$ – Simply Beautiful Art Oct 8 '17 at 14:10

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