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Joint Entrance Examination
Mathematics
Definite Integration
Question 222
Question 222 of 246
Medium
MCQ
The value of \[\lim _\limits{n \rightarrow \infty} \sum_\limits{k=1}^n \frac{n^3}{\left(n^2+k^2\right)\left(n^2+3 k^2\right)}\] is :
A
\[\frac{\pi}{8(2 \sqrt{3}+3)}\]
B
\[\frac{(2 \sqrt{3}+3) \pi}{24}\]
C
\[\frac{13 \pi}{8(4 \sqrt{3}+3)}\]
D
\[\frac{13(2 \sqrt{3}-3) \pi}{8}\]
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Official solution
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\begin{aligned} & \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)} \end{aligned}
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\] \[=\int_\limits0^1 \frac{d x}{3\left(1+x^2\right)\left(\frac{1}{3}+x^2\right)}\] \[
\begin{aligned} & =\int_\limits0^1 \frac{1}{3} \times \frac{3}{2} \frac{\left(x^2+1\right)-\left(x^2+\frac{1}{3}\right)}{\left(1+x^2\right)\left(x^2+\frac{1}{3}\right)} d x \\ & =\frac{1}{2} \int_\limits0^1\left[\frac{1}{x^2+\left(\frac{1}{\sqrt{3}}\right)^2}-\frac{1}{1+x^2}\right] d x \\ & =\frac{1}{2}\left[\sqrt{3} \tan ^{-1}(\sqrt{3} x)\right]_0^1-\frac{1}{2}\left(\tan ^{-1} x\right)_0^1 \\ & =\frac{\sqrt{3}}{2}\left(\frac{\pi}{3}\right)-\frac{1}{2}\left(\frac{\pi}{4}\right)=\frac{\pi}{2 \sqrt{3}}-\frac{\pi}{8} \\ & =\frac{13 \pi}{8 \cdot(4 \sqrt{3}+3)} \end{aligned}
\]
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