Special Series — Question 10

Question 10 – Solution

Step 1: Decompose the gap-two denominator.

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

Step 2: Telescope the finite sum.

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

Step 3: Take the limit.

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

Step 4: Find the exact remainder.

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