<s>[INST] Proving that $\frac{\pi}{4}$$=1-\frac{\eta(1)}{2}+\frac{\eta(2)}{4}-\frac{\eta(3)}{8}+\cdots$ After some calculations with WolframAlfa, it seems that 
$$
\frac{\pi}{4}=1+\sum_{k=1}^{\infty}(-1)^{k}\frac{\eta(k)}{2^{k}}
$$
Where
$$
\eta(n)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{n}}
$$
is the Dirichlet Eta function. 
Could it be proved that this is true, or false?
Thanks.

ADDED:
If we consider the Dirichlet Beta function 
$$
\beta(z)=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)^{z}}
$$
We can write this as 
$$
\beta(1)=1+\sum_{k=1}^{\infty}(-1)^{k}\frac{\eta(k)}{2^{k}}
$$

ADDED:
I recently also noted that
$$
\frac{\pi}{4}=\sum_{k=1}^{\infty}\frac{\eta(k)}{2^{k}}
$$
So summing both of the expressions 
Whe have that
$$
\frac{\pi}{2}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{2^{2k}}
$$
and 
$$
\frac{1}{2}=\sum_{k=1}^{\infty}\frac{\eta(2k-1)}{2^{2k-1}}
$$
 [/INST] If we can change the order of summation, we obtain
$$\begin{align}
1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\eta(k) &= 1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^k} \\
&= 1 + \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=1}^\infty \frac{(-1)^k}{(2n)^k}\\
&= 1 + \sum_{n=1}^\infty (-1)^{n+1} \left(-\frac{1}{2n}\right)\frac{1}{1 + \frac{1}{2n}}\\
&= 1 + \sum_{n=1}^\infty \frac{(-1)^n}{2n+1},
\end{align}$$
which is the Leibniz series for $\frac{\pi}{4}$.
The convergence of the double sum is not absolute, so the change of summation requires a justification. We obtain that by a slightly more circumspect computation:
$$\begin{align}
1 + \sum_{k=1}^\infty \frac{(-1)^k}{2^k}\eta(k) &= 1 - \frac{\eta(1)}{2} + \sum_{k=2}^\infty \frac{(-1)^k}{2^k}\eta(k)\\
&= 1 - \frac{\eta(1)}{2} + \sum_{k=2}^\infty\frac{(-1)^k}{2^k}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^k}\\
&= 1 - \frac{\eta(1)}{2} + \sum_{n=1}^\infty (-1)^{n+1}\sum_{k=2}^\infty \frac{(-1)^k}{(2n)^k}\\
&= 1 - \frac{\eta(1)}{2} + \sum_{n=1}^\infty (-1)^{n+1}\left(-\frac{1}{2n}\right)^2\frac{1}{1+\frac{1}{2n}}\\
&= 1 - \frac{\eta(1)}{2} + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n(2n+1)}\\
&= 1 - \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n} + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n(2n+1)}\\
&= 1 + \sum_{n=1}^\infty (-1)^{n+1}\left(\frac{1}{2n(2n+1)} -\frac{1}{2n}\right)\\
&= 1 + \sum_{n=1}^\infty \frac{(-1)^n}{2n+1}.
\end{align}$$
Here the change of order of summation is unproblematic, since
$$\sum_{n=1}^\infty\sum_{k=2}^\infty \frac{1}{(2n)^k} = \sum_{n=1}^\infty \frac{1}{2n(2n-1)} < \infty.$$
</s>