Series - The Basics — Question 5

*Question 5 Consider $\displaystyle\sum_{n=1}^{\infty}\frac1{n(n+2)}$.

1. Find a partial-fraction decomposition that produces telescoping cancellation.

2. Derive the exact $N$th partial sum and verify it for $N=1,2,3$.

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

*Question 5 – Solution

Write $1/[n(n+2)]=A/n+B/(n+2)$. Multiplying by $n(n+2)$ gives $1=A(n+2)+Bn$. Matching coefficients yields $A=1/2$ and $B=-1/2$, so $$\frac1{n(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right).$$

Hence $$s_N=\frac12\left(1+\frac12-\frac1{N+1}-\frac1{N+2}\right) =\frac34-\frac1{2(N+1)}-\frac1{2(N+2)}.$$ The surviving terms are $1$, $1/2$, $-1/(N+1)$, and $-1/(N+2)$. For $N=1,2,3$, the formula gives $1/3$, $11/24$, and $21/40$, matching direct addition.

Since both remaining reciprocal terms approach zero, $$S=\frac34.$$

Subtracting the partial sum from $S$ gives the exact positive tail $$R_N=S-s_N=\frac1{2(N+1)}+\frac1{2(N+2)}.$$