# How to prove $\sum_{i=1}^n \frac1{4i^2-1}=\frac{1}{2}-\frac{1}{2 \times (2n+1)}$?

How could we show $$\sum_{i=1}^n \frac{1}{4i^2-1}=\frac{n}{2n+1}=\frac{1}{2}-\frac{1}{2 \times (2n+1)}$$ I am looking for an answer other than induction as in the actual problem I am suppose to find out a representation of the sum (I used wolfram alpha to get that representation).Experimenting with smaller sums it is not much difficult to identify the form $\frac{n}{2n+1}$ however, I would appreciate if anybody show me other methods to obtain those forms.

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– Martin Sleziak yesterday

This sum can be computed using telescoping: $\frac1{4n^2-1}=\frac12\left(\frac1{2n-1}-\frac1{2n+1}\right)$, so in the sum all terms but $\frac12\cdot 1$ and $-\frac12\cdot\frac1{2n+1}$ cancel.

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Ehhh I was very close to this solution.Thanks anyways :-) – Quixotic Aug 5 '11 at 11:06

It may be worth trying $$\frac{1}{{4n^2 - 1}} = \frac{1}{{(2n + 1)(2n - 1)}} = \frac{1}{{2(2n - 1)}} - \frac{1}{{2(2n + 1)}}$$ and considering telescoping series.

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