TITLE: Proof $1 -\frac{1}{5} + \frac{1}{9} - \frac{1}{13} + ... = \frac{\pi + 2\ln(1+\sqrt2)}{4\sqrt2}$
QUESTION [7 upvotes]: I'm trying to show that
$$1 -\frac{1}{5} + \frac{1}{9} - \frac{1}{13} + \cdots = \frac{\pi + 2\ln(1+\sqrt2)}{4\sqrt2}.$$
I thought of using the power series for $\tanh^{-1}z$ which I found was $\sum_{n=0}^\infty \frac{z^{2n+1}}{2n+1}$ and playing with $z=i^{3/2}$ but I get some unwanted terms that I can't get rid of...
Can you help me please?

REPLY [2 votes]: You may use your idea too and write the expansion of :
\begin{align}
\arctan(z)&=\sum_{n=0}^\infty (-1)^n\frac{z^{2n+1}}{2n+1}\\
\operatorname{arctanh}(z)&=\sum_{n=0}^\infty \frac{z^{2n+1}}{2n+1}\\
\end{align}
so that 
$$\frac 12(\arctan(z)+\operatorname{arctanh}(z))=\sum_{n=0}^\infty \frac{z^{4n+1}}{4n+1}$$
Noting that $\;\arctan(z)=-i\,\operatorname{arctanh}(i\,z))\,$ and searching a root of $z^4=-1$ should help you to conclude... especially if you remember that $\;\operatorname{arctanh}(x)=\dfrac{\ln(1+x)-\ln(1-x)}2$.