Special Series — Question 4

Question 4 – Solution

Step 1: Decompose one summand.

Write $1/[n(n+1)]=A/n+B/(n+1)$. Multiplication by $n(n+1)$ gives $1=A(n+1)+Bn$, so $A=1$ and $B=-1$. Hence $$\frac1{n(n+1)}=\frac1n-\frac1{n+1}.$$

Step 2: Telescope a finite sum.

$$\begin{aligned} s_N&=\left(1-\frac12\right)+\left(\frac12-\frac13\right) +\cdots+\left(\frac1N-\frac1{N+1}\right)\\ &=1-\frac1{N+1}. \end{aligned}$$ All interior reciprocal terms cancel; only the two boundary terms remain.

Step 3: Pass to the limit and measure the error.

Since $1/(N+1)\to0$, the series sums to $S=1$. The exact remainder is $$R_N=S-s_N=\frac1{N+1}.$$ As a check, $s_N-s_{N-1}=1/[N(N+1)]$, the original $N$th term.