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Question #41523

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the limit: $L = \lim_{n \to \infty} \left( \frac{1}{n} \sum_{k=1}^{n-1} \sum_{m=0}^{\infty} \binom{2m}{m} \left(\frac{k}{2n}\right)^{2m} + \sum_{k=n+1}^{2n} \frac{1}{k} \sum_{m=0}^{\infty} \binom{2m}{m} \left(\frac{n}{2k}\right)^{2m} \right)$

Options

A.
$\frac{\pi}{2} + \ln(2+\sqrt{3})$
B.
$\frac{\pi}{4} + \ln(2+\sqrt{3})$
C.
$\pi + \ln(2-\sqrt{3})$
D.
$\frac{\pi}{2} + \frac{1}{2}\ln(3)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Integral as the Limit of a Sum|Binomial Theorem for Negative/Fractional Indices|Denominator Restriction

Why This Question Is Tricky

Common Mistakes

This question tests the 'Denominator Restriction' trap inherent in fractional binomial expansions.

The series $\sum \binom{2m}{m} y^m$ converges to $(1-4y)^{-1/2}$ strictly for $|y| < 1/4$ .

A common mistake is to blindly integrate such series term-by-term over an extended interval, ignoring the radius of convergence, which leads to divergent integrals.

By presenting a limit of a sum partitioned at $k=n$ (corresponding to $x=1$ ), the problem forces the student to recognize the distinct convergence domains: $x < 1$ for the first series and $x > 1$ for the second.

The student must convert the valid series into their closed-form algebraic functions before applying the limit of a sum to evaluate the resulting improper integrals.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Integral as the Limit of a Sum
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free