
Originally Posted by
Dystopia
i) $\displaystyle \sum_{r=0}^{\infty} (-1)^{r} \left(\frac{1}{4r+1} + \frac{1}{4r+3} \right)$
Note that,
$\displaystyle \sum_{n=-\infty}^{-1} (-1)^n \left( \frac{1}{4n+1}+\frac{1}{4n+3} \right) = \sum_{n=-\infty}^{-1}(-1)^{(1-n)-1}\left( \frac{-1}{4(1-n)+3}+\frac{-1}{4(1-n)+1} \right) $$\displaystyle = \sum_{n=0}^{\infty} (-1)^n \left( \frac{1}{4n+1}+\frac{1}{4n+3} \right)$
Therefore,
$\displaystyle \sum_{n=-\infty}^{\infty}(-1)^n \left( \frac1{4n+1}+\frac1{4n+3}\right) = 2\sum_{n=0}^{\infty}(-1)^n\left( \frac1{4n+1}+\frac1{4n+3}\right)$.
Now define $\displaystyle f(z) = \frac{1}{4z+1}+\frac{1}{4z+3}$ this function has poles at $\displaystyle \xi = -1/4$ and $\displaystyle \mu = -3/4$.
If $\displaystyle F(z) = \pi f(z) \csc \pi z$ then by complex analysis methods we have:
$\displaystyle \sum_{n=-\infty}^{\infty} (-1)^n f(n) = - \mbox{Res}(F,\xi) - \mbox{Res}(F,\mu) = \frac{\pi}{2\sqrt{2}}+\frac{\pi}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}}$.
Thus, $\displaystyle \sum_{n=0}^{\infty} (-1)^n \left( \frac{1}{4n+1}+\frac{1}{4n+3} \right) = \frac{1}{2}\cdot \frac{\pi}{\sqrt{2}}$.