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

Step-by-step solution

1. 1

   Decompose into partial fractions. From $\frac{1}{n(n+2)}=\frac{A}{n}+\frac{B}{n+2}$ we get $1=A(n+2)+Bn$, so $A=\frac12$ and $B=-\frac12$, giving $\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)$.

2. 2

   Notice the gap of two. Unlike a neighbouring-term telescope, each negative piece cancels the positive piece two places later. That means two terms survive at the front and two at the back, not one.

3. 3

   Write the partial sum. $S_N=\frac{1}{2}\left[\left(1+\frac12\right)-\left(\frac{1}{N+1}+\frac{1}{N+2}\right)\right]$, since only $\frac11,\frac12$ never get cancelled from the front and $\frac{1}{N+1},\frac{1}{N+2}$ never get cancelled from the back.

4. 4

   Take the limit. As $N\to\infty$ the two trailing fractions vanish, leaving $S=\frac{1}{2}\left(1+\frac12\right)$.

5. 5

   Simplify. $\frac{1}{2}\cdot\frac{3}{2}=\frac{3}{4}$.

6. 6

   Numerical check. Summing the first $2\times10^{6}$ terms gives $0.7499995$, matching $\frac34$ to six decimals.