Special Series — Question 10

*Question 10 Evaluate $\displaystyle\sum_{n=1}^{\infty}\frac1{n(n+2)}$.

1. Find the partial-fraction decomposition.

2. Display the two-step telescoping cancellation in a finite partial sum.

3. Find the sum and exact remainder after $N$ terms.

*Question 10 – Solution

Solving $1/[n(n+2)]=A/n+B/(n+2)$ gives $A=1/2$ and $B=-1/2$. Thus $$\frac1{n(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right).$$

$$\begin{aligned} s_N&=\frac12\left[\left(1+\frac12+\frac13+\cdots+\frac1N\right) -\left(\frac13+\frac14+\cdots+\frac1{N+2}\right)\right]\\ &=\frac12\left(1+\frac12-\frac1{N+1}-\frac1{N+2}\right). \end{aligned}$$ The cancellation forms two interlaced chains because the indices differ by two.

Both ending reciprocals tend to zero, so $$S=\frac12\left(1+\frac12\right)=\boxed{\frac34}.$$

$$R_N=S-s_N=\frac1{2(N+1)}+\frac1{2(N+2)}.$$ As a check, subtracting consecutive partial-sum formulas reproduces $1/[N(N+2)]$.