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.
[Math] How to prove $\sum_{i=1}^n \frac1{4i^2-1}=\frac{1}{2}-\frac{1}{2 \times (2n+1)}$
algebra-precalculussummation
Best Answer
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.