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For all $p \in \mathbb N$, I want to prove

$$\sum_{n=p+1}^\infty \frac1{n^2-p^2} = \frac1{2p} \left(1+\frac12 + \cdots + \frac1{2p} \right).$$

Up to now, I've approached the problem using induction and/or partial fractions. When I use induction, I kind of manage to show the base case for $p = 1$ (I write down "all" terms of the infinite series and we see that all but $1+\frac12$ and the factor $\frac12$ vanish), but I have no idea how to do the induction step. I've tried writing

$$\sum_{n=p+2}^\infty \frac1{n^2-p^2} = \frac1{2p} \left(1 + \frac12 + \cdots + \frac1{2p} \right) - \frac1{2p+1}, $$

but I fail to see how to simplify this. I also know that I can rewrite e.g. $$\frac1{n^2-p^2} = \frac1{2n(p+n)} - \frac1{2n(p-n)},$$

but whenever I try to simplify this term, I eventually end up with the initial fraction again...

So my question is: How do I have to approach this problem? Also, can you provide a hint which might bring me to a further point than I am right now?

Thanks in advance.

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$p$ is fixed while $n$ varies, so the partial fraction expansion you want is $\frac{1}{n^2-p^2}=\frac{1}{2p}\left ( \frac{1}{n-p}-\frac{1}{n+p} \right )$. – Chris Eagle Oct 23 '11 at 9:49
Wow, how could I have overlooked that? :( – Huy Oct 23 '11 at 10:03
@J. M.: It seems that for $p=0$ the relation is not true. – user 1618033 May 31 '12 at 11:44
@Chris: I only edited the question, but: the restriction $p \in \mathbb N$ implies that $p$ can't be zero, no? Or are you French? – J. M. May 31 '12 at 12:10
up vote 11 down vote accepted

Use partial fractions to write $$ \begin{align} \sum_{n=p+1}^N \frac{1}{n^2-p^2} &=\frac{1}{2p}\sum_{n=p+1}^N \frac{1}{n-p}-\frac{1}{n+p}\\ &=\frac{1}{2p}\sum_{n=1}^{N-p}\frac{1}{n}-\frac{1}{2p}\sum_{n=2p+1}^{N+p}\frac{1}{n}\\ &=\frac{1}{2p}\sum_{n=1}^{2p}\frac{1}{n}-\frac{1}{2p}\sum_{n=N-p+1}^{N+p}\frac{1}{n}\tag{1} \end{align} $$ Taking the limit of $(1)$ as $N\to\infty$, we get $$ \sum_{n=p+1}^\infty \frac{1}{n^2-p^2}=\frac{1}{2p}\sum_{n=1}^{2p}\frac{1}{n}\tag{2} $$

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We have for $N\geq 2p$: \begin{align*} \sum_{n=p+1}^{N+p}\frac 1{n^2-p^2}&=\sum_{j=1}^{N}\frac 1{(j+p)^2-p^2}\\ &=\sum_{j=1}^{N}\frac 1{j^2+2jp}\\ &=\frac 1{2p}\sum_{j=1}^{N}\frac {j+2p-j}{j(j+2p)}\\ &=\frac 1{2p}\sum_{j=1}^{N}\frac 1j-\frac 1{j+2p}\\ &=\frac 1{2p}\left(\sum_{j=1}^{2p}\frac 1j+\sum_{j=2p+1}^N\frac 1j-\sum_{j=2p+1}^{N+2p}\frac 1j\right)\\ &=\frac 1{2p}\left(\sum_{j=1}^{2p}\frac 1j-\sum_{j=N+1}^{N+2p}\frac 1j\right)\\ \sum_{n=p+1}^{N+p}\frac 1{n^2-p^2}&=\frac 1{2p}\sum_{j=1}^{2p}\frac 1j-\frac 1{2p}\sum_{k=1}^{2p}\frac 1{k+N}. \end{align*} We get the result taking the limit $N\to \infty$.

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Consider the following form of the sum without loss of generality when the initial sum tends to $\infty$: $$\frac{1}{2p}\sum_{t=0}^\infty\left(\sum_{n=(2t+1)p+1}^{(2t+3)p}\frac{1}{n-p}-\sum_{n=(2t+1)p+1}^{(2t+3)p}\frac{1}{n+p}\right)$$

We obtain that the last sum goes to $0$, and the other sums excepting the first one cancel. Therefore, we get that:

$$\sum_{n=p+1}^\infty \frac{1}{n^2-p^2}=\frac{1}{2p}\sum_{n=p+1}^{3p}\frac{1}{n-p}=\frac{1}{2p}\sum_{n=1}^{2p}\frac{1}{n}$$ The key observation is to group the sums in 2p terms sums that cancel. So, what i typed above may be simply written as: $$\frac{1}{2p} [(1+\frac{1}{2}+...+\frac1{2p}) - (\frac{1}{2p+1}+\frac1{2p+2}+...+\frac1{4p}) + (\frac{1}{2p+1}+\frac1{2p+2}+...+\frac1{4p}) - (\frac1{4p+1}+\frac1{4p+2}+...+\frac1{6p}) + (\frac1{4p+1}+\frac1{4p+2}+...+\frac1{6p}) - ... ]$$ The proof is complete.

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