JEE Main 2022 (Online) 30th July Evening ShiftQuestion 1
$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{n(n^2+k^2)}{(n^4+n^2k^2+k^4)}$ is equal to
Definite Integration
Solution
Concept. Divide top and bottom by $n^4$ to get a Riemann sum in $r=\frac kn$; then use $z=x-\frac1x$. Formulas used. (1) $\lim_{n\to\infty}\frac1n\sum_{k=1}^ng\big(\frac kn\big)=\int_0^1g(x)\,dx$(2) $\dfrac{1+x^2}{1+x^2+x^4}=\dfrac{1+\frac1{x^2}}{\big(x-\frac1x\big)^2+3}$Step 1 - Riemann sum. $$\frac{n(n^2+k^2)}{n^4+n^2k^2+k^4}=\frac1n\cdot\frac{1+r^2}{1+r^2+r^4},\quad r=\frac kn.$$ By (1) the limit is $\int_0^1\frac{1+x^2}{1+x^2+x^4}dx$. Step 2 - Use (2). $z=x-\frac1x$, $dz=\big(1+\frac1{x^2}\big)dx$, $z:-\infty\to0$: $$\int_{-\infty}^0\frac{dz}{z^2+3}=\frac1{\sqrt3}\Big(0+\frac\pi2\Big)=\frac{\pi}{2\sqrt3}.$$ Answer: (C)