JEE Main 2024 — 30 January, Shift 1 — Question 6

$\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{n^{3}}{n^{4}\left(1+\frac{k^{2}}{n^{2}}\right)\left(1+\frac{3 k^{2}}{n^{2}}\right)}$

$=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \frac{n^{3}}{\left(1+\frac{k^{2}}{n^{2}}\right)\left(1+\frac{3 k^{2}}{n^{2}}\right)}$

$=\int_{0}^{1} \frac{d x}{3\left(1+x^{2}\right)\left(\frac{1}{3}+x^{2}\right)}$

$=\int_{0}^{1} \frac{1}{3} \times \frac{3}{2} \frac{\left(x^{2}+1\right)-\left(x^{2}+\frac{1}{3}\right)}{\left(1+x^{2}\right)\left(x^{2}+\frac{1}{3}\right)} d x$

$=\frac{1}{2} \int_{0}^{1}\left[\frac{1}{x^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}}-\frac{1}{1+x^{2}}\right] d x$

$=\frac{1}{2}\left[\sqrt{3} \tan ^{-1}(\sqrt{3} x)\right]_{0}^{1}-\frac{1}{2}\left(\tan ^{-1} x\right)_{0}^{1}$

$=\frac{\sqrt{3}}{2}\left(\frac{\pi}{3}\right)-\frac{1}{2}\left(\frac{\pi}{4}\right)=\frac{\pi}{2 \sqrt{3}}-\frac{\pi}{8}$

$=\frac{13 \pi}{8 .(4 \sqrt{3}+3)}$