Question

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

is equal to :

• A. ${\tan }^{-1}\left(2\right)$
• B. ${\tan }^{-1}\left(3\right)$
• C. $\pi /4$
• D. $\pi /2$

Answer 1

The correct option is A ${\tan }^{-1}\left(2\right)$

$\underset{n\to \infty }{lim}(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+...+\frac{1}{5n})$
$=\underset{n\to \infty }{lim}(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+...+\frac{n}{n^2+(2n{)}^2})$
$=\underset{n\to \infty }{lim}\sum _{r=1}^{2n}\frac{n}{n^2+r^2}$
$=\underset{n\to \infty }{lim}\sum _{r=1}^{2n}\frac{1}{n}\left(\frac{1}{1+{\left(\frac{r}{n}\right)}^2}\right)$
$=\underset{0}{\overset{2}{\int }}\frac{dx}{1+x^2}={\tan }^{-1}\left(2\right)$