Q60 / 120 · VITEEE 2010 · Mathematics
$\lim _{\mathrm{n} \rightarrow \infty}\left[\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\ldots .+\frac{1}{n(n+1)}\right]$ is equal to
- Correct answerA1
- B$-1$
- C0
- DNone of these
Solution
$$
\begin{array}{l}
\lim _{n \rightarrow \infty}\left[\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots . .+\frac{1}{n(n+1)}\right] \\
=\lim _{n \rightarrow \infty}\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots \ldots\right. \\
=\lim _{n \rightarrow \infty}\left(1-\frac{1}{n+1}\right)=\lim _{n \rightarrow \infty} \frac{n}{n+1} \\
=\lim _{n \rightarrow \infty} \frac{n}{n\left(1+\frac{1}{n}\right)}=\lim _{n \rightarrow \infty} \frac{1}{\left(1+\frac{1}{n}\right)}=1
\end{array}
$$
Hence, the correct answer is option: 'A'.