The value of n ( k=1^n n^3(n^2+k^2)(n^2+3k^2)) is: — Solution | JEE Mathematics

MCQMediumJEE 2024Definite Integrals

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

Answer

Correct answer:B

Standard Method

Given:

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

Find: The value of $L$ .

Rewrite the summand by factoring $n^2$ from each denominator term:

\frac{n^3}{(n^2+k^2)(n^2+3k^2)}=\frac{1}{n\left(1+\left(\frac{k}{n}\right)^2\right)\left(1+3\left(\frac{k}{n}\right)^2\right)}

Hence,

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

This is a Riemann sum for

\int_0^1 \frac{dx}{(1+x^2)(1+3x^2)}

Now use partial fractions:

\frac{1}{(1+x^2)(1+3x^2)}=\frac{-\frac12}{1+x^2}+\frac{\frac32}{1+3x^2}

Therefore,

L=-\frac12\int_0^1\frac{dx}{1+x^2}+\frac32\int_0^1\frac{dx}{1+3x^2}

Evaluate the integrals:

\int_0^1\frac{dx}{1+x^2}=\tan^{-1}(1)-\tan^{-1}(0)=\frac{\pi}{4}

and

\int_0^1\frac{dx}{1+3x^2}=\frac{1}{\sqrt{3}}\tan^{-1}(\sqrt{3}x)\Big|_0^1=\frac{1}{\sqrt{3}}\cdot\frac{\pi}{3}

So,

L=-\frac12\cdot\frac{\pi}{4}+\frac32\cdot\frac{\pi}{3\sqrt{3}}=-\frac{\pi}{8}+\frac{\pi}{2\sqrt{3}}

L=\frac{\pi(4-\sqrt{3})}{8\sqrt{3}}

Rationalizing,

L=\frac{\pi(4\sqrt{3}-3)}{24}=\frac{13\pi}{8(4\sqrt{3}+3)}

Therefore, the correct option is B.

Integral Identification

The key observation is to write the sum in the form

\sum_{k=1}^{n}\frac{1}{n}f\left(\frac{k}{n}\right)

with

f(x)=\frac{1}{(1+x^2)(1+3x^2)}

Then, as $n\to\infty$ ,

\sum_{k=1}^{n}\frac{1}{n}f\left(\frac{k}{n}\right)\to \int_0^1 f(x)\,dx

This converts the limit directly into a definite integral, after which standard integration gives the required value matching option B.

Writing the summand incorrectly after dividing by $n^2$ . The numerator becomes part of a factor $\frac{1}{n}$ , not $1$ . Always factor $n^2$ from both denominator brackets carefully before converting to a Riemann sum.
Using the wrong interval for the Riemann sum. Since $\frac{k}{n}$ runs from values near $0$ to $1$ , the integral is over $[0,1]$ , not $[1,n]$ or $[0,\infty)$ .](streamdown:incomplete-link)
Making an error in the partial fraction decomposition. The form must be $\frac{A}{1+x^2}+\frac{B}{1+3x^2}$ and the constants should satisfy the identity exactly before integrating.

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