The value of $\lim _{n \rightarrow \infty} \left( \sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!} \right)$ is:

  • A
    $\frac{4}{3}$
  • B
    $2$
  • C
    $\frac{7}{3}$
  • D
    $\frac{5}{3}$

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