Given a positive odd integer $n$, compute $\sum_{m\neq n, m \text{ odd}} \frac1{n^2-m^2}$ 
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. 
 A: 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 
$$\begin{align*}
&\frac1{4n}\left(\sum_{j=0}^{k-1}\frac1{k+j+1}+\sum_{j\ge 2k+1}\left(\frac1{k-j}+\frac1{k+j+1}\right)+\sum_{j=k+1}^{2k}\frac1{k+j+1}\right)\\
&\qquad=\frac1{4n}\left(\left(\sum_{j\ge 0}\frac1{k+j+1}-\frac1{2k+1}\right)+\sum_{j\ge 2k+1}\frac1{k-j}\right)\\
&\qquad=\frac1{4n}\left(\sum_{i\ge k+1}\frac1i-\frac1n-\sum_{i\ge k+1}\frac1i\right)\\
&\qquad=-\frac1{4n^2}\;.
\end{align*}$$
