Question

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

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

Answer 1

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

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

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

$=\underset{n\to \infty }{lim}\frac{1}{2}(\frac{3-1}{1.3}+\frac{5-3}{3.5}+...\frac{(2n+1)-(2n-1)}{(2n+1)(2n-1)})$

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

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

$=\frac{1}{2}(1-0)=\frac{1}{2}$