# Finding a Derivation for a particular Series

This series came up when working out a physics problem, and I have been unable to derive the sum with any rigor. Here is the series and it's probable evaluation...

$$\sum_{odd m\neq n}^{\infty}\frac{1}{n^2-m^2}=\frac{-1}{4n^2}$$

To clarify, the sum is being taken over all odd $m$ except the specific term where $m$ is equal to $n$. $n$ is being held constant at an arbitrary odd integer. I arrived at the result somewhat empirically after looking at the first couple $n$ and seeing that it is a collapsing sum. I hope that someone can give me a more justifiable derivation, and/or check/correct my result.

• It's a small thing, but since $n,m$ have definite parity it may be easier to take $n=2k+1$ and $m=2l+1$. That gets rid of the odd constraint (and the factor of four, if I'm eyeballing it right.) Commented Oct 4, 2014 at 2:59
• Also, I imagine you mean $m \neq \pm n$ for the summation. Commented Oct 4, 2014 at 3:11

Hint: $\displaystyle\sum_{m=\text{odd}}\frac1{m^2-x^2}=\sum_{k=0}^\infty\frac1{(2k+1)^2-x^2}=\frac\pi4\cdot\frac{\tan\bigg(\dfrac\pi2x\bigg)}x.~$ Then your sum becomes
$\displaystyle\lim_{x\to n}\bigg[\frac\pi4\cdot\frac{\tan\bigg(\dfrac\pi2x\bigg)}x-\frac1{n^2-x^2}\bigg]$, where odd $n=2p+1$.
• @TylerHG: Using the fact that $\tan\bigg(\dfrac\pi2-a\bigg)=\cot a$, employ the Taylor series of $\cot a$ for $a\to0$, where $a=\dfrac\pi2(n-x)$. This is made possible by the periodicity of the $($co$)$tangent function. Commented Oct 5, 2014 at 1:48