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Question #41879

Difficulty: 8/10 | 🟢 Free

The Question

Let $L = \lim_{n \to \infty} n^2 \left( \frac{2n^2 + 3n + 1}{6n^2} - \sum_{r=1}^n \sin\left(\frac{r^2}{n^3}\right) \right)$ . Find the value of $L$ .

Options

A.
$\frac{1}{42}$
B.
$0$
C.
$\frac{1}{21}$
D.
$\frac{1}{6}$

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

Sigma Notation Properties|Sum to n terms vs. General Term|Trigonometric Standard Limits

Why This Question Is Tricky

Common Mistakes

This question tests the danger of applying point-wise limits (like $\sin x \approx x$ ) inside a summation with a variable number of terms before evaluating the limit of the entire sum.

Students who prematurely approximate $\sin(r^2/n^3)$ as $r^2/n^3$ will incorrectly cancel the terms and obtain a limit of 0.

The correct approach requires identifying the general term, applying the Taylor expansion to capture the leading non-zero term, and then evaluating the limit of the sum using Faulhaber's formula or Riemann integration.

This highlights the non-distributive nature of the limit operator over an infinite sum.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sequences and Series - Class 11
Concept	Sigma Notation Properties
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free