This is obviously easy to solve with the Taylor series formula, but I wanted to solve it with contour integration. $$ \sum^\infty_{n=0}\left(\frac{1}{2}\right)^n=2. $$ So what I did is this $$ \begin{split} \sum^\infty_{n=0}\left(\frac{1}{2}\right)^n & = \pi \sum^\infty_{n=0}\left[ \operatorname{Res}\left(\frac{1}{2^z \tan( \pi z)} ,\ n\right)\right]\\ & = \pi \oint_C \frac{1}{2^z \tan( \pi z)} \, dz \end{split} $$
So $$ \begin{split} \pi \oint _{C}\frac{1}{2^z \tan( \pi z)} \, dz &= \pi \left[\int_A +\int_B +\int_C\right]\\ &= \pi \left[\int_A +\int_B +i\right] \\ \\ \therefore \sum^\infty_{n=0}\left(\frac{1}{2}\right)^n &= \pi \left[\int_A +\int_B +i\right] \end{split} $$
From the looks of things the integral across $A$ should be zero. But on wolfram alpha, the integral across $b$ diverges! So that's a real problem! I'm not sure if this is a good contour for this contour integral, but I have to include all of the poles on the right side of the reals (including zero though).
So any help solving the infinite sum with contour integration would be greatly appreciated! Thanks in advance!