JEE Main 2022 (Online) 24th June Evening ShiftQuestion 1
$\lim_{n\to\infty}\left(\frac{n^{2}}{\left(n^{2}+1\right)(n+1)}+\frac{n^{2}}{\left(n^{2}+4\right)(n+2)}+\frac{n^{2}}{\left(n^{2}+9\right)(n+3)}+\ldots+\frac{n^{2}}{\left(n^{2}+n^{2}\right)(n+n)}\right)$ is equal to
Limits & Continuity
Solution
Concept. Write the sum as a Riemann sum and integrate by partial fractions. Formulas used. (1) $\lim\frac1n\sum_{r=1}^n\phi\big(\frac rn\big)=\int_0^1\phi(x)\,dx$(2) $\frac{1}{(1+x)(1+x^2)}=\frac12\Big[\frac{1}{1+x}+\frac{1-x}{1+x^2}\Big]$Step 1 - Riemann form. The $r$-th term is $\frac{n^2}{(n^2+r^2)(n+r)}=\frac1n\cdot\frac{1}{(1+t^2)(1+t)}$, $t=\frac rn$. By (1): $$L=\int_0^1\frac{dx}{(1+x^2)(1+x)}$$ Step 2 - Partial fractions. $\frac{A}{1+x}+\frac{Bx+C}{1+x^2}$: $x=-1$ gives $A=\frac12$; comparing $x^2$ and constants gives $B=-\frac12$, $C=\frac12$. This is (2). Step 3 - Integrate. $$\int_0^1\frac{dx}{1+x}=\ln2,\quad\int_0^1\frac{1-x}{1+x^2}dx=\frac\pi4-\frac12\ln2$$ Step 4. $L=\frac12\big(\ln2+\frac\pi4-\frac12\ln2\big)=\frac\pi8+\frac14\ln2$. Answer: A