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I am trying to find the limit of an infinite series given as $$\sum\frac{1}{n^2-1}.$$ I came across the following general term of the sequence of partial sums $$3/4-\left(\frac{1}{2n}-\frac{1}{2(n+1)}\right).$$ I would appreciate assistance to understand how this expression is arrived at. I have tried breaking down the original expression into partial fractions, but cannot get to the given result.

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We have $$\begin{align}\sum_{k=2}^{n}\frac{1}{k^2-1}&=\frac 12\sum_{k=2}^{n}\left(\frac {1}{k-1}-\frac{1}{k+1}\right)\\&=\frac 12\left\{\left(\frac 11-\frac{1}{3}\right)+\left(\frac 12-\frac 14\right)+\left(\frac 13-\frac 15\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\right\}\\&=\frac 12\left(\frac 11+\frac 12-\frac 1n-\frac{1}{n+1}\right)\end{align}$$

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Hint

We have

$$\frac1{n^2-1}=\frac1{(n-1)(n+1)}\\=\frac12\left(\frac1{n-1}-\frac1{n+1}\right)=\frac12\left(\frac1{n-1}-\frac1n\right)+\frac12\left(\frac1n-\frac1{n+1}\right)$$ and now telescope. Can you take it from here?

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  • $\begingroup$ Thanks @sami this is useful $\endgroup$ – Zilore Mumba Aug 31 '14 at 19:35
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You have $$ \frac{1}{n^2-1} = \frac 1 2 \left(\ \underbrace{\frac{1}{n-1}-\frac{1}{n+1}\ }_{\text{Call this $\{$A$\}$}} \right) $$ Consequently \begin{align} & \frac12\left( \left(\frac{1}{2-1} - \frac{1}{2+1}\right) + \left(\frac{1}{3-1} - \frac{1}{3+1}\right) + \left( \frac{1}{4-1} - \frac{1}{4+1} \right) + \cdots + \{\text{A}\} \right) \\[10pt] = {} & \frac 1 2 \left( \left(\frac 1 1 - \frac 1 3\right) + \left( \frac 1 2 - \frac 1 4 \right) + \left( \frac 1 3 - \frac 1 5 \right) + \left( \frac 1 4 - \frac 1 6 \right) + \left( \frac 1 5 - \frac 1 7 \right) + \cdots +\{\text{A}\} \right). \end{align} So $+1/3$ cancels $-1/3$; $+1/4$ cancels $-1/4$; $+1/5$ cancels $-1/5$, and so on. Only a few terms at the beginning and the end do not cancel, and you're left with an expression for the partial sum whose complexity does not grow with the number of terms.

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$$\frac1{n^2-1}=\frac1{(n-1)(n+1)}=\frac12\left(\frac1{n-1}-\frac1{n+1}\right)\implies$$

$$\sum_{n=2}^M\frac1{n^2-1}=\frac12\left(1-\frac13+\frac12-\frac14+\frac13-\frac15+\ldots\frac1{M-1}-\frac1{M+1}\right)=$$

$$=\frac12\left(1+\frac12-\frac1M-\frac1{M+1}\right)\xrightarrow[M\to\infty]{}\ldots?$$

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  • $\begingroup$ There is also an uncancelled term $-\frac 1M$ $\endgroup$ – Mark Bennet Aug 31 '14 at 12:51

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