Question

$\underset{n\to \infty }{\lim }\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\dots \dots +\frac{2}{n}\right]=$

• A. $\frac{\pi }{4}+\frac{3}{2}\ln 2$
• B. $\frac{\pi }{4}+\frac{3}{4}\ln 2$
• C. $\frac{\pi }{4}-\frac{3}{2}\ln 2$
• D. $\frac{\pi }{4}+\frac{1}{2}\ln 2$

Answer 1

Answer: (A)

We have
$\underset{n\to \infty }{\lim }\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\dots \frac{2}{n}\right]$
$\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=1}^n\left(\frac{1+\frac{3r}{n}}{1+{\left(\frac{r}{n}\right)}^2}\right)=\underset{0}{\overset{1}{\int }}\left(\frac{1+3x}{1+x^2}\right)dx$
$=\underset{0}{\overset{1}{\int }}1\left(\frac{1}{1+x^2}+\frac{3x}{1+x^2}\right)dx$
$={\left[{\tan }^{-1}x+\frac{3}{2}\log \left(1+x^2\right)\right]}_0^1$
$={\tan }^{-1}1+\frac{3}{2}\log 2=\frac{\pi }{4}+\frac{3}{2}\log 2$