Question

$\underset{n\to \infty }{\lim }\left(\frac{n^2}{\left(n^2+1\right)(n+1)}+\frac{n^2}{\left(n^2+4\right)(n+2)}+\frac{n^2}{\left(n^2+9\right)(n+3)}+\dots +\frac{n^2}{\left(n^2+n^2\right)(n+n)}\right)$ is equal to

• A. $\frac{\pi }{8}+\frac{1}{4}{\log }_{e}2$
• B. $\frac{\pi }{4}+\frac{1}{8}{\log }_{e}2$
• C. $\frac{\pi }{4}-\frac{1}{8}{\log }_{e}2$
• D. $\frac{\pi }{8}+{\log }_e\sqrt{2}$

Answer 1

Answer: (A)

$\underset{n\to \infty }{\lim }\left(\sum _{r=1}^n\frac{n^2}{\left(n^2+r^2\right)(n+r)}\right)$
$=\underset{n\to \infty }{\lim }\left(\sum _{r=1}^n\frac{1}{n\left(1+{\left(\frac{r}{n}\right)}^2\right)\left(1+\left(\frac{r}{n}\right)\right)}\right]$
$=\underset{0}{\overset{1}{\int }}\frac{dx}{\left(1+x^2\right)(1+x)}=\frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{1-x}{1+x^2}dx+\frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{1}{1+x}dx$
$=\frac{1}{2}\int \left(\frac{1}{1+x^2}-\frac{x}{1+x^2}\right)dx+\frac{1}{2}(\ln (1+x){)}_0^1$
$=\frac{1}{2}{\left[\tan {}^{-1}x-\frac{1}{2}\ell n\left(1+x^2\right)\right]}_0^1+\frac{1}{2}\ell n2$
$=\frac{1}{2}\left[\frac{\pi }{4}-\frac{1}{2}\ell n2\right]+\frac{1}{2}\ell n2$
$=\frac{\pi }{8}+\frac{1}{4}\ell n2$