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

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!

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up vote 6 down vote accepted

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.

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How did you know to choose that function. Which concept is this? BTW excellent answer. – Henry Jun 8 '14 at 20:31
The series is almost geometric, but there is the term $4n+1$ on the denominator. It can be "cancelled" by differentiation of a power of $x$. – Davide Giraudo Jun 8 '14 at 20:34
OK, but what do you mean when you say 'since the question is tagged "power-series"'? I tagged power series because I thought it had something to do with G.P. – Henry Jun 8 '14 at 20:41
I introduced a function which is expressed as a power series. – Davide Giraudo Jun 8 '14 at 20:45
Thanks! I got it now. – Henry Jun 8 '14 at 20:46

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.

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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.

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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}

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