Question

Evaluate the following :
$\underset{n\to \infty }{lim}\left(\frac{1}{1.2}+\frac{1}{2.3}+\dots +\frac{1}{(n-1)n}\right)$

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

Answer 1

The correct option is A $1$

Let the nth term of the series be given by $\frac{1}{(n-1)\left(n\right)}$

This can be written as $\frac{n-(n-1)}{(n-1)\left(n\right)}$

Or, $\frac{1}{n-1}-\frac{1}{n}$

Expanding the series,

$li{m}_{n\to \infty }(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}........\frac{1}{n-1}-\frac{1}{n})$

Cancelling the terms, the expression reduces to

$li{m}_{n\to \infty }1-\frac{1}{n}$

$=1$ $[\because n\to \infty \Rightarrow \frac{1}{n}\to 0]$