The value of lim _{n rightarrow infty} left( | vedclass

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(D) We want to evaluate $\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!}$.First,rewrite the numerator: $k^3+6 k^2+11 k+5 = (

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$\frac{5}{3}$

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(D) We want to evaluate $\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!}$.First,rewrite the numerator: $k^3+6 k^2+11 k+5 = (k+1)(k+2)(k+3) - 1$.Substituting this into the sum,we get:$\sum_{k=1}^n \frac{(k+1)(k+2)(k+3)-1}{(k+3)!} = \sum_{k=1}^n \left( \frac{(k+1)(k+2)(k+3)}{(k+3)!} - \frac{1}{(k+3)!} \right)$.Simplifying the first term: $\frac{(k+1)(k+2)(k+3)}{(k+3)!} = \frac{1}{k!}$.So the sum becomes $\sum_{k=1}^n \left( \frac{1}{k!} - \frac{1}{(k+3)!} \right)$.Expanding the sum:$\left( \frac{1}{1!} - \frac{1}{4!} \right) + \left( \frac{1}{2!} - \frac{1}{5!} \right) + \left( \frac{1}{3!} - \frac{1}{6!} \right) + \dots + \left( \frac{1}{n!} - \frac{1}{(n+3)!} \right)$.As $n \rightarrow \infty$,the terms $\frac{1}{(n+1)!}, \frac{1}{(n+2)!}, \frac{1}{(n+3)!}$ approach $0$.The remaining terms are $\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} = 1 + \frac{1}{2} + \frac{1}{6} = \frac{6+3+1}{6} = \frac{10}{6} = \frac{5}{3}$.

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:

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