Proof $1 -\frac{1}{5} + \frac{1}{9} - \frac{1}{13} + ... = \frac{\pi + 2\ln(1+\sqrt2)}{4\sqrt2}$

Question

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?

Answer 1

HINT:

$$\sum_{n=0}^\infty\frac{(-1)^n}{4n+1}=\sum_{n=0}^\infty\int_0^1(-1)^nx^{4n}dx=\int_0^1\frac{dx}{1+x^4}$$

To evaluate this integral, see, for example, this answer.

Answer 2

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 $\;\operatorname{arctanh}(x)=\dfrac{\ln(1+x)-\ln(1-x)}2$.