Question

The value of
\[ \lim_{n\to\infty}\left[\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{n(n+1)}\right] \] is equal to

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

Hint:

Series of form \(\sum \frac{1}{k(k+1)}\) always telescopes to \(1-\frac{1}{n+1}\). Limit becomes 1.

Answer 1

The Correct Option is A

Step 1: Use partial fraction decomposition.
\[ \frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1} \] Step 2: Rewrite the series.
\[ \sum_{k=1}^{n}\frac{1}{k(k+1)} = \sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right) \] Step 3: Observe telescoping cancellation.
\[ \left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right) \] All middle terms cancel, leaving:
\[ 1-\frac{1}{n+1} \] Step 4: Take limit as \(n\to\infty\).
\[ \lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)=1 \] Final Answer: \[ \boxed{1} \]