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I've been playing with series involving odd values of the zeta function. Some time ago I found the following closed form $$ \sum_{k=1}^{\infty}\frac{\eta(2k+1)}{2^{2k+1}}=\frac{1}{2}-\ln(2) $$ and recently I stumbled upon this one $$ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{\zeta(2n+1)}{2^{2n+1}} $$ but had no luck, yet, finding a nice closed form. When I say "nice", I mean in terms of $ln$, $\pi$, etc.

So, maybe someone here can find and share the hypothetical nice closed form.

Thanks.

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    $\begingroup$ By switching the order of summation I get $\gamma+\Re \psi(1+i/2)$. $\endgroup$ – Olivier Oloa May 24 '16 at 22:07
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    $\begingroup$ how do you find the result for the first series ? $\endgroup$ – reuns May 24 '16 at 22:10
  • $\begingroup$ I'm trying to run away from $\psi$... $\endgroup$ – Neves May 24 '16 at 22:11
  • $\begingroup$ If it weren't alternating, I think $\eta(n)=(1-2^{1-n})\zeta(n)$ would've taken care of most of the problem. $\endgroup$ – Simply Beautiful Art May 24 '16 at 22:32
  • $\begingroup$ @user1952009, see this previous question of mine and from the answer accepted you can get to "the result for the first series". $\endgroup$ – Neves May 26 '16 at 15:42
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This is not a complete answer.

$$\sum_{n=1}^\infty(-1)^{n+1}\frac{\zeta(2n+1)}{2^{2n+1}}=\sum_{n=1}^\infty(-1)^{n+1}\frac{\sum_{k=1}^\infty\frac1{k^{2n+1}}}{2^{2n+1}}$$

$$=\sum_{k=1}^\infty\sum_{n=1}^\infty(-1)^{n+1}\frac{\frac1{k^{2n+1}}}{2^{2n+1}}$$

$$=\sum_{k=1}^\infty\sum_{n=1}^\infty(-1)^{n+1}\frac1{(2k)^{2n+1}}$$

$$=\sum_{k=1}^\infty\sum_{n=1}^\infty(-1)^{n+1}\frac1{2k(4k^2)^n}$$

$$=\sum_{k=1}^\infty\frac{-1}{2k}\sum_{n=1}^\infty(-1)^n\left(\frac1{4k^2}\right)^n$$

$$=\sum_{k=1}^\infty\frac{-1}{2k}\frac1{1+4k^2}$$

$$=-\frac12\sum_{k=1}^\infty\frac1{k(1+4k^2)}$$

According to Wolframalpha

$$\sum_{n=1}^\infty(-1)^{n+1}\frac{\zeta(2n+1)}{2^{2n+1}}=\frac12\left(2\gamma+\psi^{(0)}(1-\frac i2)+\psi^{(0)}(1+\frac i2)\right)$$

And looking at the integral representation,

$$=\gamma+\Re\psi^{(0)}(1+\frac i2)$$

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We can conclude the Simple Art's answer observing that $$\sum_{k\geq1}\frac{1}{k\left(4k^{2}+1\right)}=\sum_{k\geq1}\frac{1}{k\left(2k+i\right)\left(2k-i\right)} $$ $$\frac{i}{4}\sum_{k\geq1}\frac{1}{k\left(k+\frac{i}{2}\right)}-\frac{i}{4}\sum_{k\geq1}\frac{1}{k\left(k-\frac{i}{2}\right)} $$ $$=\frac{1}{2}\left(2\gamma+\psi^{\left(0\right)}\left(1+\frac{i}{2}\right)+\psi^{\left(0\right)}\left(1-\frac{i}{2}\right)\right) $$ hence $$\sum_{n\geq1}\left(-1\right)^{n+1}\frac{\zeta\left(2n+2\right)}{2^{2n+1}}=-\frac{1}{4}\left(2\gamma+\psi^{\left(0\right)}\left(1+\frac{i}{2}\right)+\psi^{\left(0\right)}\left(1-\frac{i}{2}\right)\right). $$

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    $\begingroup$ Hmmm... you ended up just at my starting place. I asked for something in terms of $\ln(2)$, $\pi$, etc. See my question. $\endgroup$ – Neves May 26 '16 at 6:33
  • $\begingroup$ @Neves Sorry, I didn't see the "nice" request in your question. Personally I doubt that it is possible to what do you want, and Mathematica seems agree with me. $\endgroup$ – Marco Cantarini May 26 '16 at 7:37
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    $\begingroup$ @Neves : you didn't answer to my comment above, but you are criticizing the answers ? $\endgroup$ – reuns May 26 '16 at 8:32
  • $\begingroup$ @ user1952009, I think that Marco wonderstood, my comment, because his answer was not in terms of $\ln$, etc... which I asked for explicitly. $\endgroup$ – Neves May 26 '16 at 15:46
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    $\begingroup$ the trick is to compute $\displaystyle F(a) = \sum_{k=1}^\infty \frac{\eta(k)}{a^k}$ then $\displaystyle \sum_{k=1}^\infty \frac{\eta(2k)}{a^{2k}} = \frac{F(a)+F(-a)}{2}$. but here $a$ is imaginary, and this case is not treated in your link... @Neves $\endgroup$ – reuns May 26 '16 at 15:56
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In order to clarify the four cases - and possibly correction in the question - we have:

$$ \sum_{n=1}^{\infty}\frac{\eta(2n+1)}{2^{2n+1}} \quad = \frac{1}{2}\int_{0}^{\infty}\frac{\cosh(x/2)-1}{e^{x}+1}\,dx = \frac{1}{2}\left(1-\log2\right) \tag{1} \\[6mm] $$ $$ \sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{2^{2n+1}} \quad = \frac{1}{2}\int_{0}^{\infty}\frac{\cosh(x/2)-1}{e^{x}-1}\,dx = \frac{1}{2}\left(2\log2-1\right) \tag{2} \\[6mm] $$ $$ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{\eta(2n+1)}{2^{2n+1}} = \frac{1}{2}\int_{0}^{\infty}\frac{1-\cos(x/2)}{e^{x}+1}\,dx = \frac{1}{2}\log2\,+ \qquad\quad\qquad \frac{1}{8}\left(H_{-1/2+i/4}+H_{-1/2-i/4}\right)-\frac{1}{8}\left(H_{+i/4}+H_{-i/4}\right) \tag{3} \\[6mm] $$ $$ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{\zeta(2n+1)}{2^{2n+1}} = \frac{1}{2}\int_{0}^{\infty}\frac{1-\cos(x/2)}{e^{x}-1}\,dx = \frac{1}{4}\left(H_{+i/2}+H_{-i/2}\right) \tag{4} \\[8mm] $$ $$ \small \color{blue}{H_s=\gamma+\psi(s+1)=\int_{0}^{1}\frac{1-x^s}{1-x}\,dx}\quad\text{"Generalized Harmonic Number"} $$


$$ \#\space\sum_{n=1}^{\infty}\frac{\eta(2n+1)}{2^{2n+1}} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{\Gamma(2n+1)\,\eta(2n+1)}{\Gamma(2n+1)\,2^{2n}} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{2^{2n}\,(2n)!}\int_{0}^{\infty}\frac{x^{2n}}{e^x+1}\,dx = \\ \frac{1}{2}\int_{0}^{\infty}\frac{1}{e^x+1}\left(\sum_{n=1}^{\infty}\frac{(x/2)^{2n}}{(2n)!}\right)\,dx = \frac{1}{2}\int_{0}^{\infty}\frac{\cosh(x/2)-1}{e^{x}+1}\,dx = \color{red}{\frac{1}{2}\left(1-\log2\right)} \\[8mm] $$

$$ \#\space\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\zeta(2n+1)}{2^{2n+1}} = \frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\Gamma(2n+1)\,\zeta(2n+1)}{\Gamma(2n+1)\,2^{2n}} = \\ \frac{1}{2}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{2^{2n}\,(2n)!}\int_{0}^{\infty}\frac{x^{2n}}{e^x-1}\,dx = \frac{1}{2}\int_{0}^{\infty}\frac{1}{e^x-1}\left(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(x/2)^{2n}}{(2n)!}\right)\,dx = \\ \frac{1}{2}\int_{0}^{\infty}\frac{1-\cos(x/2)}{e^{x}-1}\,dx = \color{red}{\frac{1}{4}\left(H_{+i/2}+H_{-i/2}\right)} = \frac{1}{2}\,\Re(H_{i/2}) = \frac{1}{2}\left[\gamma+\Re\left(\psi(1+i/2)\right)\right] \\[8mm] $$ And, unfortunately, neither $Re\left\{H_{i/2}\right\}$ nor $Re\left\{\psi(1+i/2)\right\}$ have known closed form in term of log, exp or similar functions.

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