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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 $\frac{3}{4} - \frac{1}{2n}-\frac{1}{2(n+1)}$?

Rondo
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  • 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 $\frac{3}{4} - \frac{1}{2n}-\frac{1}{2(n+1)}$? – Rondo Jun 09 '19 at 19:38
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    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 $$\sum_{n=2}^{N}\frac1{n^2-1}=\frac{3}{4}-\frac1{2N}-\frac{1}{2(N+1)}?$$ – clathratus Jun 09 '19 at 19:43
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    I have updated my question again. This time hopefully it makes sense. – Rondo Jun 09 '19 at 19:46
  • It makes perfect sense now. A proof is on ts way ;) – clathratus Jun 09 '19 at 19:46

3 Answers3

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Hint: $\frac{1}{n^2-1}=\frac{1}{2}(\frac{1}{n-1}-\frac{1}{n+1})$

Paul
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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 $$\begin{align} f(N)&=\frac34+\frac12 H_{N-1}-\frac12H_{N+1}\\ &=\frac34+\frac12 \int_0^1\frac{(t^{N-1}-1)-(t^{N+1}-1)}{t-1}dt\\ &=\frac34-\frac12 \int_0^1\frac{t^{N+1}-t^{N-1}}{t-1}dt\\ &=\frac34-\frac12 \int_0^1t^{N-1}\frac{t^2-1}{t-1}dt\\ &=\frac34-\frac12 \int_0^1t^{N-1}\frac{(t-1)(t+1)}{t-1}dt\\ &=\frac34-\frac12 \int_0^1t^{N-1}(t+1)dt\\ &=\frac34-\frac12 \int_0^1t^N dt-\frac12\int_0^1 t^{N-1}dt\\ &=\frac34-\frac1{2(N+1)}-\frac1{2N}. \end{align}$$

clathratus
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This series can be converted into a telescoping series. Using partial fraction decomposition, we can see that $$\frac{1}{n^2-1}=\frac{1}{(n+1)(n-1)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$$ Therefore, $$\begin{align} 2\sum_{r=2}^n\frac{1}{r^2-1}&=\sum_{r=2}^n\left(\frac{1}{r-1}-\frac{1}{r+1}\right)\\ &=\frac{1}{1}-\frac{1}{3}\\ &~+\frac{1}{2}-\frac{1}{4}\\ &~+\frac{1}{3}-\frac{1}{5}\\ &~~~~~~~\vdots~~~~~~~\vdots\\ &~+\frac{1}{n-2}-\frac{1}{n}\\ &~+\frac{1}{n-1}-\frac{1}{n+1}\\ \end{align}$$ And all the terms will cancel apart from $4$ of them, so we are left with $$2\sum_{r=2}^n\frac{1}{r^2-1}=1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}=\frac{3}{2}-\frac{1}{n}-\frac{1}{n+1}$$ which means that $$\sum_{r=2}^n\frac{1}{r^2-1}=\frac{3}{4}-\frac{1}{2n}-\frac{1}{2(n+1)}$$

We can take the limit as $n\to\infty$ to find that $$\sum_{r=1}^\infty\frac{1}{r^2-1}=\lim_{n\to\infty}\sum_{r=1}^n\frac{1}{r^2-1}=\frac{3}{4}$$

A-Level Student
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