Question

$\lim \limits_{n \rightarrow \infty}\left(\frac{n^{2}}{\left(n^{2}+1\right)(n+1)}+\frac{n^{2}}{\left(n^{2}+4\right)(n+2)}+\frac{n^{2}}{\left(n^{2}+9\right)(n+3)}+\ldots+\frac{n^{2}}{\left(n^{2}+n^{2}\right)(n+n)}\right)$ 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)

$\displaystyle\lim _{n \rightarrow \infty}\left(\displaystyle\sum_{r=1}^{n} \frac{n^{2}}{\left(n^{2}+r^{2}\right)(n+r)}\right)$
$=\displaystyle\lim _{n \rightarrow \infty}\left(\displaystyle\sum_{r=1}^{n} \frac{1}{n\left(1+\left(\frac{r}{n}\right)^{2}\right)\left(1+\left(\frac{r}{n}\right)\right)}\right]$
$=\int\limits_{0}^{1} \frac{d x}{\left(1+x^{2}\right)(1+x)}=\frac{1}{2} \int\limits_{0}^{1} \frac{1-x}{1+x^{2}} d x+\frac{1}{2} \int\limits_{0}^{1} \frac{1}{1+x} d x$
$=\frac{1}{2} \int\left(\frac{1}{1+x^{2}}-\frac{x}{1+x^{2}}\right) d x+\frac{1}{2}(\ln (1+x))_{0}^{1}$
$=\frac{1}{2}\left[\tan { }^{-1} x-\frac{1}{2} \ell n\left(1+x^{2}\right)\right]_{0}^{1}+\frac{1}{2} \ell n 2$
$=\frac{1}{2}\left[\frac{\pi}{4}-\frac{1}{2} \ell n 2\right]+\frac{1}{2} \ell n 2$
$=\frac{\pi}{8}+\frac{1}{4} \ell n 2$