Functions Question 306
Question: The value of $ \underset{n\to \infty }{\mathop{\lim }},\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+…+\frac{1}{(2n-1)(2n+1)} $ is equal to
[DCE 2005]
Options:
A) ½
B) 1/3
C) ¼
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{n\to \infty }{\mathop{\lim }},\frac{1}{2}[ ( 1-\frac{1}{3} )+( \frac{1}{3}-\frac{1}{5} )+( \frac{1}{5}-\frac{1}{7} )+…. . $ $ . +( \frac{1}{(2n-1)}-\frac{1}{(2n+1)} ) ] $ $ =\underset{n\to \infty }{\mathop{\lim }},\frac{1}{2}[ 1-\frac{1}{2n+1} ]=\frac{1}{2} $ .
BETA
