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Like user71317 in his question I am struggling to understand how we arrive at the general formula of the partial sums of a series. In my case the following series:

$$\sum_{n=2}^{\infty} \frac{1}{n^2-1}$$

The general formula of the partial sums is:

$$s_n=\sum_{i=2}^{n} \frac{1}{i^2-1} = \frac{3}{4} - \frac{1}{2n}-\frac{1}{2(n+1)}$$

I understand the answer to the linked question where the answer is the formula for the sum of a finite geometric series. Since my series is not geometric; how do I go from the series to the partial sums in this case?

Update:
How is $\sum_{i=2}^{n} \frac{1}{i^2-1}$ equal to $\frac{3}{4} - \frac{1}{2n}-\frac{1}{2(n+1)}$?

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  • $\begingroup$ Apologies, I have updated my original question. I don't fully understand how $\sum_{n=2}^{\infty} \frac{1}{n^2-1}$ is equal to $\frac{3}{4} - \frac{1}{2n}-\frac{1}{2(n+1)}$? $\endgroup$ – Rondo Jun 9 at 19:38
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    $\begingroup$ It makes no sense to say that $$\frac1{2^2-1}+\frac1{3^2-1}+\frac1{4^2-1}+\dots=\frac{3}{4}-\frac1{2n}-\frac1{2(n+1)}.$$ Perhaps you mean to ask for a proof that $$\sum_{n=2}^{N}\frac1{n^2-1}=\frac{3}{4}-\frac1{2N}-\frac{1}{2(N+1)}?$$ $\endgroup$ – clathratus Jun 9 at 19:43
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    $\begingroup$ I have updated my question again. This time hopefully it makes sense. $\endgroup$ – Rondo Jun 9 at 19:46
  • $\begingroup$ It makes perfect sense now. A proof is on ts way ;) $\endgroup$ – clathratus Jun 9 at 19:46
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If you are trying to find $$f(N)=\sum_{n=2}^{N}\frac1{n^2-1}$$ then you should note that $$\frac1{n^2-1}=\frac12\cdot\frac{1}{n-1}-\frac12\cdot\frac1{n+1}$$ so $$\begin{align} f(N)&=\frac12\sum_{n=2}^{N}\frac1{n-1}-\frac12\sum_{n=2}^{N}\frac{1}{n+1}\\ &=\frac12\sum_{n=1}^{N-1}\frac1{n}-\frac12\sum_{n=3}^{N+1}\frac1n\\ &=\frac12\sum_{n=1}^{N-1}\frac1{n}-\frac12\left[-1-\frac12+\sum_{n=1}^{N+1}\frac1n\right]\\ &=\frac34+\frac12\sum_{n=1}^{N-1}\frac1{n}-\frac12\sum_{n=1}^{N+1}\frac1n\ . \end{align}$$ Then recall that $$H_k=\sum_{i=1}^{k}\frac1i=\int_0^1\frac{t^k-1}{t-1}dt\ .$$ So we have that $$\begin{align} f(N)&=\frac34+\frac12 H_{N-1}-\frac12H_{N+1}\\ &=\frac34+\frac12 \int_0^1\frac{(t^{N-1}-1)-(t^{N+1}-1)}{t-1}dt\\ &=\frac34-\frac12 \int_0^1\frac{t^{N+1}-t^{N-1}}{t-1}dt\\ &=\frac34-\frac12 \int_0^1t^{N-1}\frac{t^2-1}{t-1}dt\\ &=\frac34-\frac12 \int_0^1t^{N-1}\frac{(t-1)(t+1)}{t-1}dt\\ &=\frac34-\frac12 \int_0^1t^{N-1}(t+1)dt\\ &=\frac34-\frac12 \int_0^1t^N dt-\frac12\int_0^1 t^{N-1}dt\\ &=\frac34-\frac1{2(N+1)}-\frac1{2N}. \end{align}$$

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  • $\begingroup$ You don't actually need the integrals in the last set of equations, since we know $H_{n+1} = H_{n-1} + \frac{1}{n} + \frac{1}{n+1}$. $\endgroup$ – Jair Taylor Jun 9 at 20:08
  • $\begingroup$ @JairTaylor Yeah, I know, but I just thought it would be a little more fun to use the integrals. $\endgroup$ – clathratus Jun 9 at 20:10
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Hint: $\frac{1}{n^2-1}=\frac{1}{2}(\frac{1}{n-1}-\frac{1}{n+1})$

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