Question

$\underset{x\to \infty }{lim}(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n(n+1)})$ is equal to

• A. $1$
• B. $-1$
• C. $0$
• D. none of these

Answer 1

Answer: (A)

$1$

let $S_n=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n(n+1)}$

$S_n=\frac{2.1}{1.2}+\frac{3.2}{2.3}+\frac{4.3}{3.4}+....+\frac{n+1-n}{n(n+1)}$

$S_n=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}$

$\therefore S_n=1-\frac{1}{n+1}$

$\underset{n\to \infty }{lim}{S}_n=1-\frac{1}{\infty }$

$=1$