Question

$\underset{n\to \infty }{lim}\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{\left(2n-1\left)\right(2n+1\right)}\right)=$

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{1}{2}$

Now,

$\underset{n\to \infty }{lim}\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{\left(2n-1\left)\right(2n+1\right)}\right)$

$=\underset{n\to \infty }{lim}\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+....+(\frac{1}{2n-1}-\frac{1}{2n+1})]$

$=\frac{1}{2}\underset{n\to \infty }{lim}[1-\frac{1}{2n+1}]$

$=\frac{1}{2}$. [ Since $\underset{n\to \infty }{lim}\frac{1}{2n+1}=0$]