Find the sum of the infinite series whose nth term is 1/(4n^2 - 1), summed from n = 1 to infinity.

Step-by-step solution

1. 1

   Factor the denominator. $4n^2-1$ is a difference of squares: $(2n-1)(2n+1)$. Recognising a product of two linear factors is the signal that partial fractions will telescope.

2. 2

   Split into partial fractions. Write $\frac{1}{(2n-1)(2n+1)}=\frac{A}{2n-1}+\frac{B}{2n+1}$. Clearing denominators gives $1=A(2n+1)+B(2n-1)$; setting $n=\frac12$ gives $A=\frac12$ and $n=-\frac12$ gives $B=-\frac12$.

3. 3

   Write the general term. $\frac{1}{4n^2-1}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$.

4. 4

   Form the partial sum. $S_N=\frac{1}{2}\left[\left(1-\frac13\right)+\left(\frac13-\frac15\right)+\cdots+\left(\frac{1}{2N-1}-\frac{1}{2N+1}\right)\right]$. Every interior fraction appears once positive and once negative, so only the first and last survive: $S_N=\frac{1}{2}\left(1-\frac{1}{2N+1}\right)$.

5. 5

   Take the limit. As $N\to\infty$, $\frac{1}{2N+1}\to 0$, so $S_N\to\frac12$.

6. 6

   Numerical check. Adding the first $2\times10^{6}$ terms gives $0.49999987$, consistent with the closed form $S_N=\frac12-\frac{1}{2(2N+1)}$.