I was trying to solve this series and I have an exam in a week. I can't understand how to find its sum although I managed to rework it by transforming $\frac{1}{m^2 - n^2}$ into $\frac{1}{2m}(\frac{1}{m+n} - \frac{1}{n-m})$.
I'm sure there is a way, as my book explicitly confirms this as being convergent and having a sum.
Any help is appreciated.
I get $-\frac{3}{4m^2}$.
\begin{align} s(m) &=\sum_{\underset{m \neq n}{n = 1}}^{\infty} \frac{1}{m^2 - n^2}\\ &=\sum_{\underset{m \neq n}{n = 1}}^{\infty} \frac{1}{2m}\left(\frac{1}{m+n} + \frac{1}{m-n}\right)\\ &= \frac{1}{2m}\sum_{\underset{m \neq n}{n = 1}}^{\infty}\left(\frac{1}{m+n} + \frac{1}{m-n}\right)\\ &= \frac{1}{2m}\left(\sum_{n = 1}^{m-1}\left(\frac{1}{m+n} + \frac{1}{m-n}\right) +\sum_{n = m+1}^{\infty}\left(\frac{1}{m+n} + \frac{1}{m-n}\right)\right)\\ &= \frac{1}{2m}(t(m)+u(m)) \end{align}
\begin{align} t(m) &=\sum_{n = 1}^{m-1}\left(\frac{1}{m+n} +\frac{1}{m-n}\right)\\ &= \sum_{n = 1}^{m-1}\frac{1}{m+n} +\sum_{n = 1}^{m-1}\frac{1}{m-n}\\ &= \sum_{n = m+1}^{2m-1}\frac{1}{n} + \sum_{n = 1}^{m-1}\frac{1}{n}\\ &=H_{2m-1}-H_m+H_{m-1}\\ &=H_{2m-1}-\dfrac1{m}\\ u(m) &=\sum_{n = m+1}^{\infty}\left(\frac{1}{m+n} + \frac{1}{m-n}\right)\\ &=\lim_{k \to \infty}\sum_{n = m+1}^{k}(\frac{1}{m+n} + \frac{1}{m-n})\\ &=\lim_{k \to \infty}\left(\sum_{n = m+1}^{k}\frac{1}{m+n} +\sum_{n = m+1}^{k}\frac{1}{m-n}\right)\\ &=\lim_{k \to \infty}\left(\sum_{n = m+1}^{k}\frac{1}{m+n} -\sum_{n = m+1}^{k}\frac{1}{n-m}\right)\\ &=\lim_{k \to \infty}\left(\sum_{n = 2m+1}^{k+m}\frac{1}{n} - \sum_{n = 1}^{k-m}\frac{1}{n}\right)\\ &=\lim_{k \to \infty}\left(\sum_{n = 2m+1}^{k-m}\frac{1}{n}+\sum_{n = k-m+1}^{k+m}\frac{1}{n} - (\sum_{n = 1}^{2m}\frac{1}{n}+\sum_{n = 2m+1}^{k-m}\frac{1}{n})\right)\\ &=\lim_{k \to \infty}\left(\sum_{n = k-m+1}^{k+m}\frac{1}{n} - \sum_{n = 1}^{2m}\frac{1}{n}\right)\\ &=-H_{2m} \qquad\text{since }\sum_{n = k-m+1}^{k+m}\frac{1}{n}<\frac{2m}{k-m+1} \to 0\\ \end{align} So,
\begin{align} s(m) &=\frac1{2m}(H_{2m-1}-\dfrac1{m}-H_{2m})\\ &=\frac1{2m}(-\dfrac1{m}-\dfrac1{2m})\\ &=-\frac{3}{4m^2}\\ \end{align}
We have the series were $n<m$ $\frac 1{2m}(\frac {1}{m + 1} + \frac {1}{m + 2} +\frac {1}{m + 3}+\cdots+ \frac {1}{2m - 1}- (\frac {1}{1-m} + \frac {1}{2-m} +\frac {1}{3-m}+\cdots+ \frac {1}{-1}))$
$\frac 1{2m}(\frac {1}{m + 1} + \frac {1}{m + 2} +\frac {1}{m + 3}+\cdots+ \frac {1}{2m - 1} + \frac {1}{1} + \frac {1}{2} +\frac {1}{3}+\cdots+ \frac {1}{m-1}))$
Then we skip $n = m$
And we go from $m<n<2m$
$\frac 1{2m}(\frac {1}{2m + 1} + \frac {1}{2m + 2} +\frac {1}{2m + 3}+\cdots+ \frac {1}{3m - 1}- (\frac {1}{1} + \frac {1}{2} +\frac {1}{3}+\cdots+ \frac {1}{m-1}))$
and we have some stuff that cancels with the first partial sum
$\frac 1{2m}(\frac {1}{m + 1} + \frac {1}{m + 2} +\frac {1}{m + 3}+\cdots+ \frac {1}{2m - 1} + \frac {1}{2m + 1} + \frac {1}{2m + 2} +\frac {1}{2m + 3}+\cdots+ \frac {1}{3m - 1})$
Note that we have skipped $\frac {1}{2m}$
$n\ge 2m$
$\frac 1{2m}(\frac {1}{3m} - \frac {1}{m} + \frac {1}{3m+1} - \frac {1}{m+1}\cdots)$
Now we are subtracting every term from $\frac {1}{m+1}$ on that we added since the beginning.
$\frac {1}{2m}(-\frac {1}{m}-\frac {1}{2m}) = -\frac {3}{4m^2}$
$$
\begin{align}
\sum_{\substack{n=1\\n\ne m}}^\infty\frac1{m^2-n^2}
&=\lim_{N\to\infty}\sum_{\substack{n=1\\n\ne m}}^N\frac1{m^2-n^2}\tag1\\
&=\lim_{N\to\infty}\frac1{2m}\sum_{\substack{n=1\\n\ne m}}^N\left(\frac1{m-n}+\frac1{m+n}\right)\tag2\\
&=\lim_{N\to\infty}\frac1{2m}\sum_{\substack{n=-N\\n\ne0,n\ne m,n\ne-m}}^N\frac1{m+n}\tag3\\
&=\lim_{N\to\infty}\frac1{2m}\sum_{\substack{n=m-N\\n\ne m,n\ne 2m,n\ne0}}^{m+N}\frac1n\tag4\\
&=\lim_{N\to\infty}\frac1{2m}\left(-\frac1m-\frac1{2m}+\sum_{\substack{n=m-N\\n\ne0}}^{m+N}\frac1n\right)\tag5\\[6pt]
&=\lim_{N\to\infty}\frac1{2m}\left(-\frac1m-\frac1{2m}+\sum_{n=N-m+1}^{m+N}\frac1n\right)\tag6\\[12pt]
&=\frac1{2m}\left(-\frac1m-\frac1{2m}\right)\tag7\\[12pt]
&=-\frac3{4m^2}\tag8
\end{align}
$$
Explanation:
$(1)$: write infinite sum as a limit
$(2)$: use partial fractions
$(3)$: write as a single sum
$(4)$: substitute $n\mapsto n-m$
$(5)$: move the missing terms out front
$(6)$: cancel terms with their negatives
$(7)$: take the limit using $\sum\limits_{n=N-m+1}^{m+N}\frac1n\le\frac{2m}{N-m+1}$
$(8)$: simplify
To show the convergence, we can see that the tail series is $$\sum_{n=1}^{\infty}\frac{1}{m^{2}-(2m+n)^{2}}=-\sum_{n=1}^{\infty}\frac{1}{3m^{2}+4mn+n^{2}},$$ and since $n^{2}\leq 3m^{2}+4mn+n^{2},$ this is bounded absolutely by $\sum_{n=1}^{\infty}1/n^{2}$.
Using this same idea, observe that if $n=m+k,$ $$\frac{1}{2m}\left(\frac{1}{m+n}-\frac{1}{n-m}\right)=\frac{1}{2m}\left(\frac{1}{2m+k}-\frac{1}{k}\right),$$ so if we take the sum over $k\geq1,$ we obtain $$\sum_{n\geq m+1}\frac{1}{m^{2}-n^{2}}=-\frac{1}{2m}\sum_{k=1}^{2m}\frac{1}{k},$$ since the remaining terms telescope. The sum of the first $m-1$ terms is $$\sum_{n=1}^{m-1}\frac{1}{2m}\left(\frac{1}{m+n}+\frac{1}{m-n}\right)=\frac{1}{2m}\left(\sum_{n=m+1}^{2m-1}\frac{1}{n}+\sum_{n=1}^{m-1}\frac{1}{n}\right)=\frac{1}{2m}\left(\sum_{k=1}^{2m-1}\frac{1}{k}-\frac{1}{m}\right).$$
Then the whole sum equals $$-\frac{1}{2m}\cdot\frac{1}{2m}-\frac{1}{2m}\cdot\frac{1}{m}=-\frac{3}{4m^{2}}.$$
