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JEE Mains · Maths · STD 11 - 12. limits

\(\lim _{n \rightarrow \infty}\left(\sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!}\right)\) = __________

  1. A \(4 / 3\)
  2. B 2
  3. C \(7 / 3\)
  4. D \(5 / 3\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(5 / 3\)

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Detailed explanation

\(\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!} \) \( =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+6-1}{(k+3)!} \) \( =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{(k+1)(k+2)(k+3)-1}{(k+3)!} \)…
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