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.

  • $\begingroup$ 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.) $\endgroup$ Commented Oct 4, 2014 at 2:59
  • $\begingroup$ Also, I imagine you mean $m \neq \pm n$ for the summation. $\endgroup$ Commented Oct 4, 2014 at 3:11

1 Answer 1


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 $n=2p+1$.

  • $\begingroup$ Try differentiating the natural logarithm of Euler's infinite product formula for the cosine function. $\endgroup$
    – Lucian
    Commented Oct 4, 2014 at 3:25
  • $\begingroup$ Very nice indeed, thank you! This sum just melted away $\endgroup$ Commented Oct 4, 2014 at 6:01
  • $\begingroup$ I've kept playing with it but I can't seem to get the limits to work out $\endgroup$ Commented Oct 5, 2014 at 1:13
  • $\begingroup$ I actually don't think the limit exists. Could you steer me correctly if I'm wrong $\endgroup$ Commented Oct 5, 2014 at 1:23
  • 1
    $\begingroup$ @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. $\endgroup$
    – Lucian
    Commented Oct 5, 2014 at 1:48

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