<s>[INST] How to show $\sum_{k=1}^{80} {1 \over \frac{k}{81} - \frac{1}{2} - \frac{\iota}{2}} + {1 \over \frac{k}{81} - \frac{1}{2} +\frac{\iota}{2}} = 0$? I came across this sum while doing the homework my teacher gave us on series. It was originally $$a_k = \frac{k}{81}, S = \sum_{k=1}^{80} {a_k^2 \over 1+2a_k^2-2a_k}$$
I then decomposed it into partial fractions and ended up with
$40+ \sum_{k=1}^{80} {1 \over \frac{k}{81} - \frac{1}{2} - \frac{\iota}{2}} + {1 \over \frac{k}{81} - \frac{1}{2} +\frac{\iota}{2}}$. I know the answer is 40 from WolframAlpha, however I don't know how to go about showing that the sum with the complex numbers is 0.
How should I proceed?
 [/INST] There is no need to use complex numbers. Here the keyword is symmetry: note that $a_{81-k}=1-a_k$ and therefore
\begin{align}
2S&=\sum_{k=1}^{80} \frac{a_k^2}{1-2a_k(1-a_k)}+\sum_{k=1}^{80} \frac{a_{81-k}^2}{1-2a_{81-k}(1-a_{81-k})}\\
&=\sum_{k=1}^{80} \frac{a_k^2}{1-2a_k(1-a_k)}+\sum_{k=1}^{80} \frac{(1-a_k)^2 }{1-2(1-a_k)a_k}\\
&=\sum_{k=1}^{80} \underbrace{\frac{a_k^2+(1-a_k)^2}{1-2a_k(1-a_k)}}_{=1}=80
\end{align}
which implies that $S=40$.
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