A) 1
B) -1
C) 0
D) None of these
Correct Answer: A
Solution :
\[\underset{n\to \infty }{\mathop{\lim }}\,\left[ \frac{1}{1\cdot 2}+\frac{1}{1\cdot 3}+\frac{1}{3\cdot 4}+.........+\frac{1}{n\,(n+1)} \right]\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\left[ \left( 1-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+...... \right.\] \[\left. +\left( \frac{1}{n}-\frac{1}{n+1} \right) \right]\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\left( 1-\frac{1}{n+1} \right)=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{n+1}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n}{n\left( 1+\frac{1}{n} \right)}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{\left( 1+\frac{1}{n} \right)}=1\]
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