9
$\begingroup$

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 $(3)?$

$\endgroup$
  • 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
5
$\begingroup$

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 $\color{red}{(I)}$.

QED

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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.