Question

$\underset{n\to \infty }{lim}[\left(\frac{1}{1.2}\right)+\left(\frac{1}{2.3}\right)+.....+\left(\frac{1}{n.(n+1)}\right)]=$

Answer 1

We have

$[\left(\frac{1}{1.2}\right)+\left(\frac{1}{2.3}\right)+...+\left(\frac{1}{n(n+1)}\right)]$

$=[(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+....(\frac{1}{n}-\frac{1}{n+1})]=[1-\frac{1}{n+1}].....\left(1\right)$

${lim}_{n\to \infty }[\left(\frac{1}{1.2}\right)+\left(\frac{1}{2.3}\right)+...+\left(\frac{1}{n(n+1)}\right)]$

${lim}_{n\to \infty }\left[(1-\frac{1}{(n+1)})\right]$ (using (1)

$=1-0[\because {lim}_{n\to \infty }\left[(\frac{1}{(n+1)}=0)\right]]$

$=1$