In the given series there is sum of multiple of corresponding terms of two A.P.’s. The two A.P.’s are (i) $\frac{1}{1},\frac{1}{2},\frac{1}{3}$ , ... and (ii) $\frac{1}{2},\frac{1}{3},\frac{1}{4}$ , ... Now the nth term of sum is $a_n$ = (nth term of the sequence formed by first A.P.) × (nth term of the sequence formed by second A.P.) = $\frac{1}{n} \times \frac{1}{n+1}$ = $\frac{1}{n(n+1)}$ = $\frac{n+1-n}{n(n+1)}$ = $\frac{n+1}{n(n+1)} - \frac{n}{n(n+1)}$ = $\frac{1}{n} - \frac{1}{n+1}$ Hence, the sum to n terms is, $S_n$ = $\sum_{k=1}^{n} a_k$ = $\sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right)$ = $\sum_{k=1}^{n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k+1}$ = $\left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \right)$ - $\left( \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n+1} \right)$ = 1 + $\frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3}$ + ... + $\frac{1}{n} - \frac{1}{n} - \frac{1}{n+1}$ = 1 - $\frac{1}{n+1}$ = $\frac{n+1-1}{n+1}$ = $\frac{n}{n+1}$