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Question #40936

Difficulty: 8/10 | 🟢 Free

The Question

Let Sn=∑k=0n(n+k−1k)1nkS_n = \sum_{k=0}^{n} \binom{n+k-1}{k} \frac{1}{n^k}. Evaluate the limit: lim⁡n→∞(Sne)n\lim_{n \to \infty} \left( \frac{S_n}{e} \right)^n

Options

  • A.

    e\sqrt{e}

  • B.

    1e\frac{1}{\sqrt{e}}

  • C.

    ee

  • D.

    11

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

The 1^∞ Form Shortcut|Properties of Binomial Coefficients|Binomial Theorem for Negative/Fractional Indices

Why This Question Is Tricky

Common Mistakes

The question tests the evaluation of a 1∞1^\infty limit where the base is a finite sum of binomial coefficients.

The trap lies in mistakenly identifying the sum ∑k=0n(n+k−1k)n−k\sum_{k=0}^n \binom{n+k-1}{k} n^{-k} as the standard binomial expansion (1+1/n)n(1 + 1/n)^n, which leads to the incorrect limit e−1/2e^{-1/2} (Option B).

To solve correctly, students must recognize the coefficients belong to the Generalized Binomial Theorem for negative indices, specifically (1−1/n)−n(1 - 1/n)^{-n}.

Because the exponent nn approaches infinity, the finite sum must be transitioned to the infinite series expansion, with the realization that the tail remainder RnR_n is asymptotically negligible (decaying faster than 1/n1/n).

This stabilizes the indeterminate form, allowing the correct application of the 1∞1^\infty shortcut via Taylor expansion of the logarithmic limit.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterLimits and Derivatives - Class 11
ConceptThe 1^∞ Form Shortcut
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree