Question

$\displaystyle\lim_{n\to\infty} \left[\frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + .... + \frac{1}{n\left(n+1\right)}\right] $ is equal to

• A. 1
• B. -1
• C. 0
• D. None of these

Answer 1

Answer: (A)

$\displaystyle\lim_{n\to\infty} \left[\frac{1}{1.2} + \frac{1}{2.3}+ \frac{1}{3.4} +....+ \frac{1}{n\left(n+1\right)}\right] $
$ =\displaystyle\lim_{n\to\infty} \left[ \left(1-\frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) +\left(\frac{1}{3} -\frac{1}{4}\right)+.....+ \left(\frac{1}{n}- \frac{1}{n+1}\right)\right] $
$ =\displaystyle\lim_{n\to\infty} \left(1- \frac{1}{n+1}\right) =\lim_{n\to\infty} \frac{n}{n+1} $
$ =\displaystyle\lim_{n\to\infty} \frac{n}{n\left(1+ \frac{1}{n}\right)} = \lim_{n\to\infty} \frac{1}{\left(1+ \frac{1}{n}\right)} = 1 $