Possible Duplicate:
Computing $\sum_{m \neq n} \frac{1}{n^2-m^2}$
Given a positive odd integer $n$, compute $\displaystyle\sum_{m\neq n, \; \text{and} \; m \; \text{odd}} \frac1{n^2-m^2}$. If the indexing is confusing, here is the indexing set $M=\{1,3,5,\ldots\}\setminus\{n\}$ where $n$ is an odd integers greater than or equal to one. $\displaystyle\sum_{m\in M} \frac1{n^2-m^2}$. I tried breaking this into partial fractions $\frac1{n^2-m^2}=\frac1{2n}(\frac1{m+n}-\frac1{m-n})$ to no use.
Suppose that $n=2k+1$ and $m=2j+1$; then
$$\begin{align*} \frac1{n^2-m^2}&=\frac1{2n}\left(\frac1{n-m}+\frac1{n+m}\right)\\ &=\frac1{2n}\left(\frac1{2(k-j)}+\frac1{2(k+j+1)}\right)\\ &=\frac1{4n}\left(\frac1{k-j}+\frac1{k+j+1}\right)\;. \end{align*}$$
Now split the sum into the terms with $m<n$ and the terms with $m>n$ and write it as
$$\frac1{4n}\left(\sum_{j=0}^{k-1}\left(\frac1{k-j}+\frac1{k+j+1}\right)+\sum_{j>k}\left(\frac1{k-j}+\frac1{k+j+1}\right)\right)\;.$$
Now $$\sum_{j=0}^{k-1}\frac1{k-j}=\sum_{i=1}^k\frac1i\;,$$ and $$\sum_{j=k+1}^{2k}\frac1{k-j}=-\sum_{i=1}^k\frac1i\;,$$ so these terms cancel out, and we’re left with
$$\begin{align*} &\frac1{4n}\left(\sum_{j=0}^{k-1}\frac1{k+j+1}+\sum_{j\ge 2k+1}\left(\frac1{k-j}+\frac1{k+j+1}\right)+\sum_{j=k+1}^{2k}\frac1{k+j+1}\right)\\ &\qquad=\frac1{4n}\left(\left(\sum_{j\ge 0}\frac1{k+j+1}-\frac1{2k+1}\right)+\sum_{j\ge 2k+1}\frac1{k-j}\right)\\ &\qquad=\frac1{4n}\left(\sum_{i\ge k+1}\frac1i-\frac1n-\sum_{i\ge k+1}\frac1i\right)\\ &\qquad=-\frac1{4n^2}\;. \end{align*}$$
