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Mathematics Question on Definite Integral

lim⁡n→∞[11.2+12.3+13.4+....+1n(n+1)]\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

Explanation

Solution

lim⁡n→∞[11.2+12.3+13.4+....+1n(n+1)]\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] =lim⁡n→∞[(1−12)+(12−13)+(13−14)+.....+(1n−1n+1)]=\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] =lim⁡n→∞(1−1n+1)=lim⁡n→∞nn+1=\displaystyle\lim_{n\to\infty} \left(1- \frac{1}{n+1}\right) =\lim_{n\to\infty} \frac{n}{n+1} =lim⁡n→∞nn(1+1n)=lim⁡n→∞1(1+1n)=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