Alternating series with eta dirichlet

Let $\eta$ denote Dirichlet's eta function. Prove that

$$\frac{\pi}{4}=1+\sum_{k=1}^{\infty}(-1)^{k}\frac{\eta(k)}{2^{k}}$$

Solution

 \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}. \\
&= 1 +\frac{\pi}{4} -1 \\
&=\frac{\pi}{4}
\end{align*}

The change of summation is justified due to absolute convergence.