Question

The value of $\underset{n\to \infty }{lim}\left\{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)-n}\right\}$ is

Answer 1

We have,

${lim}_{n\to \infty }\{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{(n-1).n}\}$

$={lim}_{n\to \infty }\{\frac{1(2-1)}{1.2}+\frac{1(3-2)}{2.3}+\frac{1(4-3)}{3.4}+......+\frac{1(n-(n-1))}{(n-1).n}\}$

$={lim}_{n\to \infty }\{\frac{1.2}{1.2}-\frac{1.1}{1.2}+\frac{1.3}{2.3}-\frac{1.2}{2.3}+\frac{1.4}{3.4}-\frac{1.3}{3.4}+.......+\frac{n}{(n-1).n}-\frac{(n-1)}{(n-1).n}\}$

$={lim}_{n\to \infty }\{1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{n-1}-\frac{1}{n}\}$

$=1$

Hence, this is the answer.