Equality proof of function $g_N(x) = \sum_{n=-N}^N \frac{1}{x+n}$ I have a function
$$ g_N(x) = \sum_{n=-N}^N \frac{1}{x+n} $$
How can I prove that this function is odd, thus $ g_N(-x) = -g_N(x)$ ?

Fold the sum in half at the middle ($n = 0$):
$$
\begin{align}
g_N(x) = \sum_{n=-N}^{n=N} \frac{1}{x+n} &= \frac{1}{x} + \sum_{n=1}^{n=N} \left( \frac{1}{x-n} + \frac{1}{x+n} \right) \\
&= \frac{1}{x} + \sum_{n=1}^{n=N} \frac{2x}{x^2 - n^2}
\end{align}
$$
Now, you can verify directly that $g_N(-x) = -g_N(x)$.
