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

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

(A) $\frac{(2 \sqrt{3}+3) \pi}{24}$

(C) $\frac{13(2 \sqrt{3}-3) \pi}{8}$

(D) $\frac{\pi}{8(2 \sqrt{3}+3)}$

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Limits

The value of

lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n ^{3}}{( n ^{2} + k ^{2} ) ( n ^{2} + 3 k ^{2} )}

is :

Answer: (B)

Write the sum as a Riemann sum:

n \to \infty lim \frac{1}{n} k = 1 \sum n \frac{1}{( 1 + \frac{k ^{2}}{n ^{2}} ) ( 1 + \frac{3 k ^{2}}{n ^{2}} )} = \int_{0}^{1} \frac{d x}{( 1 + x ^{2} ) ( 1 + 3 x ^{2} )}

By partial fractions,

\frac{1}{( 1 + x ^{2} ) ( 1 + 3 x ^{2} )} = \frac{1}{2} [\frac{3}{1 + 3 x ^{2}} - \frac{1}{1 + x ^{2}}]

\int_{0}^{1} = \frac{1}{2} [3 tan^{- 1} (3 x) - tan^{- 1} x]_{0}^{1} = \frac{3}{2} \cdot \frac{π}{3} - \frac{1}{2} \cdot \frac{π}{4} = \frac{π}{2 3} - \frac{π}{8} = \frac{13 π}{8 ( 4 3 + 3 )}

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Single correct option