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

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.

## marked as duplicate by Guy Fsone, Community♦Jan 16 '18 at 23:29

• math.stackexchange.com/q/265277?rq=1 – Guy Fsone Jan 16 '18 at 23:20
• Sorry I didn't realize I can put latex code in the search bar. I'm gonna do it next time. Thank you everyone for the answers! – bencemeszaros Jan 16 '18 at 23:32

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.

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.

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.

• Your equivalent is wrong as written. The $\sim$ has a specific meaning, and here you're missing the constant $4$ on the RHS. – Clement C. Jan 16 '18 at 23:14
• i meant asymptotically equivalent to. what does it mean? – qbert Jan 16 '18 at 23:16
• $a_n \sim_{n\to\infty} b_n$ if $\lim_{n\to\infty}\frac{a_n}{b_n} =1$ (that's the simple definition; the actual one, equivalent when $b_n$ can't cancel, is that $a_n-b_n = o(b_n)$) – Clement C. Jan 16 '18 at 23:18
• I see, I will amend the above then – qbert Jan 16 '18 at 23:19

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