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.


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:
$$\frac{3}{2}\cdot \frac{5}{4}\cdots\frac{2N+1}{2N}\cdot\frac{1}{\sqrt{N}}\sim\frac{2N+1}{N\sqrt{\pi}}\sim \frac{2}{\sqrt{\pi}}$$
