Question

$\displaystyle\lim _{n \rightarrow \infty}\left[\frac{n+3}{n^{2}+1^{2}}+\frac{n+6}{n^{2}+2^{2}}+\frac{n+9}{n^{2}+3^{2}}+\ldots \ldots+\frac{2}{n}\right]=$

• 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
$\displaystyle\lim _{n \rightarrow \infty}\left[\frac{n+3}{n^{2}+1^{2}}+\frac{n+6}{n^{2}+2^{2}}+\frac{n+9}{n^{2}+3^{2}}+\ldots \frac{2}{n}\right]$
$\displaystyle\lim _{n \rightarrow \infty} \frac{1}{n} \displaystyle\sum_{r=1}^{n}\left(\frac{1+\frac{3 r}{n}}{1+\left(\frac{r}{n}\right)^{2}}\right)=\int\limits_{0}^{1}\left(\frac{1+3 x}{1+x^{2}}\right) d x$
$=\int\limits_{0}^{1} 1\left(\frac{1}{1+x^{2}}+\frac{3 x}{1+x^{2}}\right) d x $
$=\left[\tan ^{-1} x+\frac{3}{2} \log \left(1+x^{2}\right)\right]_{0}^{1}$
$=\tan ^{-1} 1+\frac{3}{2} \log 2=\frac{\pi}{4}+\frac{3}{2} \log 2$