$\lim_{n \rightarrow \infty}\left(\frac{n^{2}}{(n^{2}+1)(n+1)} + \frac{n^{2}}{(n^{2}+4)(n+2)} + \frac{n^{2}}{(n^{2}+9)(n+3)} + \ldots + \frac{n^{2}}{(n^{2}+n^{2})(n+n)}\right)$ is equal to

(A) $\frac{\pi}{8} + \frac{1}{4}\log_{e}2$

(B) $\frac{\pi}{4} + \frac{1}{8}\log_{e}2$

(C) $\frac{\pi}{4} - \frac{1}{8}\log_{e}2$

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

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Limits

lim_{n \to \infty} (\frac{n ^{2}}{( n ^{2} + 1 ) ( n + 1 )} + \frac{n ^{2}}{( n ^{2} + 4 ) ( n + 2 )} + \frac{n ^{2}}{( n ^{2} + 9 ) ( n + 3 )} + \dots + \frac{n ^{2}}{( n ^{2} + n ^{2} ) ( n + n )})

is equal to

Answer: (A)

The sum can be written as:

lim_{n \to \infty} \sum_{r = 1}^{n} \frac{n ^{2}}{( n ^{2} + r ^{2} ) ( n + r )} = lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n ( 1 + ( r / n ) ^{2} ) ( 1 + ( r / n ))} \cdot \frac{1}{n}

This is a Riemann sum that converges to:

\int_{0}^{1} \frac{d x}{( 1 + x ^{2} ) ( 1 + x )}

Using partial fractions:

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

Integrating:

\frac{1}{2} [tan^{- 1} x - \frac{1}{2} ln (1 + x^{2}) + ln (1 + x)]_{0}^{1}

= \frac{1}{2} [\frac{π}{4} - \frac{1}{2} ln 2 + ln 2] = \frac{π}{8} + \frac{1}{4} ln 2

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option