Question

The value of
$\underset{n\to \infty }{lim}\{\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+....+\frac{1}{(2n+1)(2n+3)}\}$
is

• A. $1$
• B. $1/2$
• C. $-1/2$
• D. None of the above

Answer 1

Answer: (B)

$1/2$

rth term of the expression is given by

$T_r=\frac{(2r+3)-(2r+1)}{2(2r+1)(2r+3)}=\frac{1}{2}(\frac{1}{2r+1}-\frac{1}{2r+3})$

Given expression will become

$\underset{n\to \infty }{lim}\frac{1}{2}(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}\cdots -\frac{1}{2n+3})=\underset{n\to \infty }{lim}\frac{1}{2}(\frac{1}{1}-\frac{1}{2n+3})=1/2$