How to Prove : $ \gamma +\ln\left(\frac{\pi}{4}\right) = \sum_{n=2}^{\infty} \frac{(-1)^{n} \zeta{(n)}}{2^{n-1}n} $ How to Prove : 
$$ \gamma +\ln\left(\frac{\pi}{4}\right) = \sum_{n=2}^{\infty} \frac{(-1)^{n} \zeta{(n)}}{2^{n-1}n} $$
I have tried looking at Series definitions of the Polygamma function from which we can obtain $\gamma$ but I'm a little bit lost since the given definitions on Wikipedia are not exactly like this one.
Thank you kindly for your help and time.
 A: 
Lemma:
Let $f(z)=\sum_{n=2}^{\infty} a_nz^n$ be convergent with radius $>1.$ Then:
$$\sum_{n=2}^{\infty} a_n\zeta(n)=\sum_{k=1}^{\infty} f\left(\frac1k\right)$$

Proof:
$$\begin{align}\sum_{n=2}^{\infty} a_n\zeta(n)&=\sum_{n=2}^{\infty} a_n\sum_{k=1}^{\infty} \frac{1}{k^n} \\
&=\sum_{k=1}^{\infty}\sum_{n=2}^{\infty}a_n\left(\frac 1k\right)^n\\
&=\sum_{k=1}^{\infty}f\left(\frac1k\right)
\end{align}$$

Now, in your case, $a_n=\frac{(-1)^{n}}{2^{n-1}n}$ gives $$f(z)=2\sum_{n=2} \frac{(-z/2)^n}{n}=z-2\log(1+z/2)$$
Now, $$\sum_{k=1}^{N}f(1/k)=H_N - 2\log\left(\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}\right)$$
Now, $H_N-\log N\to \gamma.$ So the limit is equal to the limit $$\gamma -2 \log\left(\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}\cdot\frac{1}{\sqrt{N}}\right)$$ as $N\to\infty.$
Thus, you just need to show:
$$\lim_{N\to\infty}\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}\cdot\frac{1}{\sqrt{N}}=\frac{2}{\sqrt{\pi}}$$
But: $$\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}=\frac{2N+1}{2^{2N}}\binom{2N}{N}$$
And we have that $\binom{2n}{n}\sim \frac{2^{2n}}{\sqrt{\pi n}}$ (see here.)
So we have:
$$\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}\cdot\frac{1}{\sqrt{N}}\sim\frac{2N+1}{N\sqrt{\pi}}\sim \frac{2}{\sqrt{\pi}}$$
A: The problem is readily reduced to the evaluation of well-known infinite sums and products when approaching in the following way
\begin{align*}
\sum_{n\ge2}\frac{(-1)^n\zeta(n)}{n2^{n-1}}&=2\sum_{n\ge2}\frac{(-1)^n}{n2^n}\sum_{k\ge1}\frac1{k^n}\\
&=2\sum_{k\ge1}\left[\frac1{2k}-\sum_{n\ge1}\frac{(-1)^{n-1}}n\left(\frac1{2k}\right)^n\right]\\
&=2\sum_{k\ge1}\left[\frac1{2k}-\log\left(1+\frac1{2k}\right)\right]
\end{align*}
Reorder the partial sums as
\begin{align*}
2\sum_{k=1}^m\left[\frac1{2k}-\log\left(1+\frac1{2k}\right)\right]&=\sum_{k=1}^m\frac1k-\sum_{k=1}^m\log\left(\left[\frac{2k+1}{2k}\right]^2\right)\\
&=\sum_{k=1}^m\frac1k+\log\left(\prod_{k=1}^m\left[\frac{2k}{2k+1}\right]^2\right)\\
&=\left[\sum_{k=1}^m\frac1k-\log\left(k+\frac12\right)\right]+\log\left(\frac12\prod_{k=1}^m\left[\frac{(2k)^2}{(2k-1)(2k+1)}\right]\right)
\end{align*}
Passing the limit $n\to\infty$, using a slight variation on the definition of the Euler-Mascheroni constant combined with the Wallis Product, we obtain
$$\lim_{m\to\infty}\left[\sum_{k=1}^m\frac1k-\log\left(k+\frac12\right)\right]+\log\left(\frac12\prod_{k=1}^m\left[\frac{(2k)^2}{(2k-1)(2k+1)}\right]\right)=\gamma+\log\left(\frac\pi4\right)$$
Therefore

$$\therefore~\sum_{n\ge2}\frac{(-1)^n\zeta(n)}{n2^{n-1}}~=~\gamma+\log\left(\frac\pi4\right)$$

Note that your given result is incorrect (I suppose you meant to write $\gamma-\log\left(\frac4\pi\right)$ instead). The result already follows before considering partial sums by using a product representation of the Gamma Function.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{\gamma + \ln\pars{\pi \over 4} =
\sum_{n = 2}^{\infty}{\pars{-1}^{n}\,\zeta\pars{n} \over 2^{n - 1}\, n}}:\
{\large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 2}^{\infty}
{\pars{-1}^{n}\,\zeta\pars{n} \over 2^{n - 1}\, n}} =
\sum_{n = 2}^{\infty}
{\pars{-1}^{n} \over 2^{n - 1}\, n}\
\overbrace{{1 \over \Gamma\pars{n}}\int_{0}^{\infty}{x^{n - 1} \over \expo{x} - 1}}^{\ds{\zeta\pars{n}}}\ \,\dd x
\\[5mm] = &\
2\int_{0}^{\infty}
{\sum_{n = 2}^{\infty}\pars{-x/2}^{n}/n! \over \expo{x} - 1}
\,{\dd x \over x} =
2\int_{0}^{\infty}
{\expo{-x/2} - 1 + x/2 \over \expo{x} - 1}\,{\dd x \over x}
\\[5mm] = &\
\int_{0}^{\infty}
{2\expo{-3x/2} - 2\expo{-x} + x\expo{-x} \over 1 - \expo{-x}}
\,{\dd x \over x}
\\[5mm] \stackrel{x\ =\ -\ln\pars{t}}{=}\,\,\,&
-\int_{0}^{1}{2t^{3/2} - 2t -t\ln\pars{t} \over 1 - t}
\,{\dd t \over t\ln\pars{t}}
\\[5mm] = &\
-\int_{0}^{1}{2t^{3/2} - 2t - t\ln\pars{t} \over 1 - t}
\pars{-\int_{0}^{\infty}t^{\xi - 1}\,\dd\xi}\,\dd t
\\[5mm] = &\
\int_{0}^{\infty}\int_{0}^{1}{2t^{\xi + 1/2} - 2t^{\xi} -
t^{\xi}\ln\pars{t} \over 1 - t}\,\dd t\,\dd\xi
\\[5mm] = &\
\int_{0}^{\infty}\bracks{%
2\int_{0}^{1}{1 - t^{\xi} \over 1 - t}\,\dd t -
2\int_{0}^{1}{1 - t^{\xi + 1/2} \over 1 - t}\,\dd t -
\int_{0}^{1}{t^{\xi}\ln\pars{t} \over 1 - t}\,\dd t}\dd\xi
\label{1}\tag{1}
\end{align}
Integral are evaluated as:
$$
\left\{\begin{array}{rcl}
\ds{\int_{0}^{1}{1 - t^{\xi} \over 1 - t}\,\dd t} & \ds{=} &
\ds{\Psi\pars{\xi + 1} + \gamma}
\\[1mm]
\ds{\int_{0}^{1}{1 - t^{\xi + 1/2} \over 1 - t}\,\dd t} & \ds{=} &
\ds{\Psi\pars{\xi + {3 \over 2}} + \gamma}
\\[5mm]
\ds{\int_{0}^{1}{t^{\xi}\ln\pars{t} \over 1 - t}\,\dd t} & \ds{=} &
\ds{\left.-\,\partiald{}{\mu}\int_{0}^{1}{1 - t^{\mu} \over 1 - t}\,\dd t\,
\right\vert_{\ \mu\ =\ \xi}}
\\[1mm] & = &
\ds{\left.-\,\partiald{\Psi\pars{\mu + 1}}{\mu}
\,\right\vert_{\ \mu\ =\ \xi} = -\Psi\, '\pars{\xi + 1}}
\end{array}\right.
$$
(\ref{1}) becomes:
\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 2}^{\infty}
{\pars{-1}^{n}\,\zeta\pars{n} \over 2^{n - 1}\, n}} =
\int_{0}^{\infty}\bracks{%
2\Psi\pars{\xi + 1} - 2\Psi\pars{\xi + {3 \over 2}}
+ \Psi\, '\pars{\xi + 1}}\dd\xi
\\[5mm] = &\
\left.2\ln\pars{\Gamma\pars{\xi + 1} \over \Gamma\pars{\xi + 3/2}}  + \Psi\pars{\xi + 1}\,\right\vert_{\ \xi\ =\ 0}^{\ \xi\ \to\ \infty}
\\[5mm] = &\
\underbrace{\lim_{\xi \to \infty}\bracks{%
2\ln\pars{\Gamma\pars{\xi + 1} \over \Gamma\pars{\xi + 3/2}}  + \Psi\pars{\xi + 1}}}_{\ds{=\ 0}}\ -\
\bracks{2\ln\pars{\Gamma\pars{1} \over \Gamma\pars{3/2}}  + \Psi\pars{1}}
\\[5mm] = &\
2\ln\pars{{1 \over 2}\,\root{\pi}} + \gamma =
\bbx{\gamma + \ln\pars{\pi \over 4}} \\ &
\end{align}
with
$\ds{\Gamma\pars{1} = 1\,,\ \Gamma\pars{1/2} = \root{\pi}}$ and
$\ds{\Psi\pars{1} = -\gamma}$.
