<s>[INST] Sum the following $\sum_{n=0}^{\infty} \frac {(-1)^n}{4^{4n+1}(4n+1)} $ Evaluate:
$$\sum_{n=0}^{\infty} \frac {(-1)^n}{4^{4n+1}(4n+1)} $$   

I rewrote the sum as $$\sum_{n=0}^{\infty} \frac {1}{4^{8n-7}(8n-7)} - \sum_{n=0}^{\infty} \frac {1}{4^{8n-3}(8n-3)}$$
  Now, I tried to express this as a Geometric Series and Partial Fraction but was unable to do so. I also tried to use Riemann Sum, but I don't know how to apply it here. Any help will be appreciated. Thanks!

 [/INST] The series for $\tanh^{-1}(x)$ and $\tan^{-1}(x)$ are
\begin{align}
\tanh^{-1}(x) &= \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1} \\
\tanh^{-1}(x) &= \sum_{n=0}^{\infty} \frac{(-1)^{n} \ x^{2n+1}}{2n+1}.
\end{align}
Adding the series together leads to
\begin{align}
\sum_{n=0}^{\infty} \frac{x^{4n+1}}{4n+1} &= \frac{1}{2} \left( \tanh^{-1}(x) + \tanh^{-1}(x) \right) 
\end{align}
and upon setting $x = e^{i \pi/4}/4$ the resulting series is
\begin{align}
\sum_{n=0}^{\infty} \frac{(-1)^{n}}{4^{4n+1} \ (4n+1)} = \frac{1}{2} \left( \tanh^{-1}\left(\frac{e^{i \pi/4}}{4}\right) + \tan^{-1}\left(\frac{e^{i \pi/4}}{4}\right) \right).
\end{align}
After some reductions it can be shown that the series in question is given by
\begin{align}
\sum_{n=0}^{\infty} \frac{(-1)^{n}}{4^{4n+1} \ (4n+1)} &= \frac{1}{4\sqrt{2}} \left[ \ln\left(\frac{17+4\sqrt{2}}{17-4\sqrt{2}}\right) + 2 \tan^{-1}\left(\frac{4\sqrt{2}}{15}\right) \right].
\end{align}
</s>