Question

$\underset{n\to \infty }{lim}[\frac{1}{1\times 3}+\frac{1}{3\times 5}+\frac{1}{5\times 7}+......\frac{1}{(2n-1)(2n+1)}]$

• A.
• B.
• C. 0
• D.

Answer 1

The correct option is A

we have, $\underset{n\to \infty }{lim}[\frac{1}{1\times 3}+\frac{1}{3\times 5}+\frac{1}{5\times 7}+......\frac{1}{(2n-1)(2n+1)}]$

Numbers in the denominator are consecutive odd number,

Difference between then is 2

Multiplying 2 in numerator and denominator

$\underset{n\to \infty }{lim}\frac{2}{2}[\frac{1}{1\times 3}+\frac{1}{3\times 5}+\frac{1}{5\times 7}+......\frac{1}{(2n-1)(2n+1)}]$

$\underset{n\to \infty }{lim}\left(\frac{1}{2}\right)[\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+......\frac{2}{(2n-1)(2n+1)}]$

$\underset{n\to \infty }{lim}\frac{1}{2}[\frac{3-1}{1\times 3}+\frac{5-3}{3\times 5}+\frac{7-5}{5\times 7}+......\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}]$

$\underset{n\to \infty }{lim}\frac{1}{2}[\frac{3}{1\times 3}-\frac{1}{1\times 3}+\frac{5}{3\times 5}-\frac{3}{3\times 5}+\frac{7}{5\times 7}-\frac{5}{5\times 7}+......\frac{(2n+1)}{(2n-1)(2n+1)}-\frac{(2n-1)}{(2n-1)(2n+1)}]$

$\underset{n\to \infty }{lim}\frac{1}{2}[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....\frac{1}{(2n-1)}-\frac{1}{(2n+1)}]$
$\underset{n\to \infty }{lim}\frac{1}{2}[1-\frac{1}{(2n+1)}]$
= 1/2