I'm trying to figure out what the sum of this series could be $$ \sum_{n=1,n\neq a,b}^{+\infty} \frac{n^2}{(n^2-a^2)(n^2-b^2)}\quad\text{with}\quad a,b\in\text{N}^+$$ but I've got no idea whatsoever: I tried differentianting or integrating in one of the parameters, but this made only the expression worse. I tried also rewriting it as $$ \sum_{n=1,n\neq b}^{+\infty} \frac{1}{n^2-a^2} + b^2 \sum_{n=1,n\neq a,b}^{+\infty} \frac{1}{(n^2-a^2)(n^2-b^2)} $$ and use the result of this post, but then I still don't know what to do with the second term. I consulted two different tables of series, but nothing came out.

Any idea on how to solve it?? Thank you very much!

  • 1
    $\begingroup$ I think that I have a problem somewhere. I delete and work again. $\endgroup$ Commented Sep 18, 2023 at 11:50

1 Answer 1


Using the fact that for an integer $m$ $$\sum_{n=1;n\neq m}^\infty\frac1{n^2-m^2}=\frac3{4m^2}\tag{0}$$ (please, see below)

Let's suppose that $a, b\in N$, and $a\neq b$ $$S=\sum_{n=1;n\neq a,b}\frac{n^2}{(n^2-a^2)(n^2-b^2)}=\frac1{b^2-a^2}\sum_{n=1;n\neq a,b}\left(\frac{b^2}{n^2-b^2}-\frac{a^2}{n^2-a^2}\right)$$ $$=\frac1{b^2-a^2}(b^2S_b-a^2S_a)\tag{1}$$ where $$S_b=\sum_{n=1;n\neq a,b}\frac1{n^2-b^2}=\left(\sum_{n=1;n\neq b}\frac1{n^2-b^2}\right)\,-\frac1{a^2-b^2}=\frac3{4b^2}\,-\frac1{a^2-b^2}\tag{2}$$ $$S_a=\sum_{n=1;n\neq a,b}\frac1{n^2-a^2}=\frac3{4a^2}\,-\frac1{b^2-a^2}\tag{3}$$ Using (1), (2) and (3) $$\boxed{\,\,S=\sum_{n=1;n\neq a,b}\frac{n^2}{(n^2-a^2)(n^2-b^2)}=\frac{a^2+b^2}{(b^2-a^2)^2}, \,\,a\neq b\,\,}$$

There is also an interesting case $a=b$ $$S=\sum_{n=1;n\neq a}\frac{n^2}{(n^2-a^2)^2}=\sum_{n=1;n\neq a}\frac1{(n^2-a^2)}+a^2\sum_{n=1;n\neq a}\frac1{(n^2-a^2)^2}$$ We can show (please, see below) that $$\sum_{n=1;n\neq a}\frac1{(n^2-a^2)^2}=-\,\frac{11}{16\,a^4}+\frac{\pi^2}{12\,a^2}\tag{4}$$ and, therefore, $$\boxed{\,\,S=\sum_{n=1;n\neq a}\frac{n^2}{(n^2-a^2)^2}=\frac{\pi^2}{12}+\frac1{16\,a^2}\,\,}$$

To prove $\operatorname{(0)}$, let' suppose that $a$ is not an integer, and $a=m+\epsilon, \epsilon\to 0$.

Then $$\sum_{n=1}^\infty\frac1{n^2-a^2}=\sum_{n=1;n\neq m}^\infty\frac1{n^2-a^2}+\frac1{m^2-(m+\epsilon)^2}$$ On the other hand, $$\sum_{n=1}^\infty\frac1{n^2-a^2}=-\frac\pi2\frac{\cot\pi a}a+\frac1{2a^2}$$ Leading $a\to m$ ($\epsilon \to 0$) $$\sum_{n=1;n\neq m}^\infty\frac1{n^2-m^2}-\frac1{2m\epsilon}+\frac1{4m^2}=-\frac{\pi}{2\pi m\epsilon}\big(1-\frac\epsilon{m}\big)+\frac1{2m^2}+O(\epsilon)$$ $$\Rightarrow\,\,\sum_{n=1;n\neq m}^\infty\frac1{n^2-m^2}=\frac3{4m^2}$$

To prove$\operatorname{(4)}$ we notice that for non-integer $a$ $$\sum_{n=1}^\infty\frac1{(n^2-a^2)^2}=\frac12\sum_{n=-\infty}^\infty\frac1{(n^2-a^2)^2}-\frac1{2a^4}$$ $$=\frac12\frac{\partial}{\partial (a^2)}\left(\sum_{n=-\infty}^\infty\frac1{n^2-a^2}\right)-\frac1{2a^4}=-\,\frac1{2a^4}+\frac\pi{4a^2}\left(\frac\pi{\sin^2\pi a}+\frac{\cot\pi a}a\right)$$ and then we use the same approach (leading $a$ to an integer $m$) as for the $\operatorname{(0)}$ above.

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    $\begingroup$ That's a fantastic answer, thank you so much! $\endgroup$
    – Rob Tan
    Commented Sep 17, 2023 at 19:43
  • $\begingroup$ Could you tell me where I made a mistake in my deleted answer ? Thanks and cheers :) $\endgroup$ Commented Sep 18, 2023 at 11:53
  • $\begingroup$ Thanks ! I just feel stupid (once moreà. $\endgroup$ Commented Sep 18, 2023 at 14:27

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