Question

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

• A. $0$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $1$

Answer 1

The correct option is D $1$

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

$\Rightarrow$ $\underset{n\to \infty }{lim}(\frac{2-1}{1.2}+\frac{3-2}{2.3}+........\frac{n+1-n}{(n-1)n})$

$\Rightarrow$ $\underset{n\to \infty }{lim}(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1})$

$\Rightarrow$ $\underset{n\to \infty }{lim}(1-\frac{1}{n+1})$

$\Rightarrow$ $1-0$

$\Rightarrow$ $1$