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!

 A: Since the question is tagged "power-series", define 
$$f(x):=\sum_{n=0}^{+\infty}\frac{(-1)^nx^{4n+1}}{4^{4n+1}(4n+1)}.$$
We want to compute $f(1)$. Since 
$$f'(x)=\sum_{n=0}^{+\infty}\frac{(-1)^nx^{4n}}{4^{4n+1}}=
\frac 14\sum_{n=0}^{+\infty}\left(\frac{-x^4}{4^4}\right)^n=\frac 14\frac 1{1+x^4/4^4}$$
and $f(0)=0$, we obtain that 
$$f(1)=\frac 14\int_0^1\frac{\mathrm{dx}}{1+\frac{x^4}{4^4}}.$$
This integral can be computed by noticing that for any $t$, 
$$t^4+1=(t^2-\sqrt 2 t+1)(t^2+\sqrt 2 t+1),$$
and doing a partial fraction decomposition.
A: This is $f(1)$ where $$f(x) = \sum_{n=0}^\infty \dfrac{(-1)^n}{4^{4n+1} (4n+1)} x^{4n+1}$$
Note that $f(0) = 0$, and differentiate term-by-term to get $f'(x)$ as a geometric series.  Then integrate.
A: Define $f(x) = \displaystyle\sum_{n = 0}^{\infty} \dfrac{(-1)^nx^{4n+1}}{4n+1}$. Notice that $f(\tfrac{1}{4}) = \displaystyle\sum_{n = 0}^{\infty} \dfrac{(-1)^n}{4^{4n+1}(4n+1)}$ is the sum in question. 
It is easy to see that this power series converges for $|x| < 1$. 
Thus, we can differentiate termwise to get $f'(x) = \displaystyle\sum_{n = 0}^{\infty} (-1)^nx^{4n} = \dfrac{1}{1+x^4}$ for all $|x| < 1$. 
Clearly, $f(0) = 0$, Hence, $f(\tfrac{1}{4}) = f(\tfrac{1}{4}) - f(0) = \displaystyle\int_{0}^{\tfrac{1}{4}}f'(x)\,dx = \displaystyle\int_{0}^{\tfrac{1}{4}}\dfrac{1}{1+x^4}\,dx$
This integral can be evaluated using partial fractions, but it is messy. 
A: 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}
