# 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 $g_N(-x) = -g_N(x)$ ?

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In addition to the algebraic answers, there is the simple intuition that your function consists of N + 1 poles located at -N, -N + 1, ..., 0, 1, ... N. These are all just horizontal translations of $1/x$ (a well-known odd function), symmetrically placed around 0, added together. – Kaz Apr 9 '13 at 23:13

$$-g_N(x)=-\sum_{n=-N}^N \frac{1}{x+n} = \sum_{n=-N}^N -\frac{1}{x+n} =\sum_{n=-N}^N \frac{1}{-x+n}$$ $$g_N(-x)=\sum_{n=-N}^N \frac{1}{-x+n}$$

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Hint: $$\frac{1}{-x + n} = -\frac{1}{x - n}$$

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HINT: Note that $$\sum_{n=M}^K a_n = \sum_{n=-K}^{-M} a_{-n}.$$

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+1 If I had any votes left, I think both our hints combine as a full answer – muzzlator Apr 9 '13 at 22:13
@muzzlator Yep :) – Hagen von Eitzen Apr 9 '13 at 22:16

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 $g_N(-x) = -g_N(x)$.

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