Question

$\underset{n\to \infty }{lim}[\frac{1}{n}+\frac{n}{(n+1{)}^2}+\frac{n}{(n+2{)}^2}+....+\frac{n}{(2n-1{)}^2}]$ is equal to:

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

Answer 1

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

$\underset{n\to \infty }{lim}\sum _{r=0}^{n-1}\frac{n}{(n+r{)}^2}=\underset{n\to \infty }{lim}\sum _{r=0}^{n-1}\frac{n}{n^2(1+\frac{r}{n})}=\underset{0}{\overset{1}{\int }}\frac{dx}{(1+x{)}^2}$
$=-{\left[\frac{1}{1+x}\right]}_0^1\Rightarrow -[\frac{1}{2}-1]=\frac{1}{2}$