How do I calculate the sum of this infinite series? $\sum\limits_{n=1}^{\infty}\frac{1}{4n^{2}-1}$

Question

Could someone please help me with how do I calculate the sum of the $$\sum_{n=1}^{\infty}\frac{1}{4n^{2}-1}$$ infinite series? I see that $$\lim_{n\rightarrow\infty}\frac{1}{4n^{2}-1}=0$$ so the series is convergent based on the Cauchy's convergence test. But how do I calculate the sum? Thank you.

Answer 1

Hint. By a fraction decomposition, one gets $$ \frac{2}{4n^{2}-1}=\frac{1}{2n-1}-\frac{1}{2n+1} $$ then one may use a telescoping sum.

Answer 2

Since$$\frac1{4n^2-1}=\frac1{(2n-1)(2n+1)}=\frac12\left(\frac1{2n-1}-\frac1{2n+1}\right),$$your series is a telescopic series.

Answer 3

$$\frac{1}{4n^2-1}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$

Answer 4

Your justification for the convergence to he series needs work (having a limit of zero of the summand does not imply the sum converges!), it does however converge a priori by noting that $$ \frac{1}{4n^2-1}=O\left(\frac{1}{n^2}\right) $$ So it converges by the $p$-test.

You may also use the partial fraction decomposition noted in the other answers and compute the telescoping series, showing that it converges.