The value of lim x ( k = 1 n k 3 + 6 k 2 + 11 k + 5 ( k + 3 ) ! ) is : [2025]

Question

The value of $\lim_{n \to \infty} (\sum_{k = 1}^{n} \frac{k^{3} + 6 k^{2} + 11 k + 5}{(k + 3)!})$ is : [2025]

Smart Achievers · Accepted Answer

(1)

$\lim_{n \to \infty} [\sum_{k = 1}^{n} \frac{k^{3} + 6 k^{2} + 11 k + 5}{(k + 3)!}]$

$= \lim_{n \to \infty} [\sum_{k = 1}^{n} \frac{k^{3} + 6 k^{2} + 11 k + 6 - 1}{(k + 3)!}]$

$= \lim_{n \to \infty} [\sum_{k = 1}^{n} \frac{(k + 3) (k + 2) (k + 1) - 1}{(k + 3)!}]$

$= \lim_{n \to \infty} [\sum_{k = 1}^{n} (\frac{1}{k!} - \frac{1}{(k + 3)!})]$

$= \lim_{n \to \infty} [\frac{1}{1!} - \frac{1}{4!} + \frac{1}{2!} - \frac{1}{5!} + \frac{1}{3!} - \frac{1}{6!} + \frac{1}{4!} - \frac{1}{7!} + . . . + \frac{1}{n!} - \frac{1}{(n + 3)!}]$

$= 1 + \frac{1}{2!} + \frac{1}{3!} = \frac{5}{3}$ .

Solution

Q.
The value of $\lim_{n \to \infty} (\sum_{k = 1}^{n} \frac{k^{3} + 6 k^{2} + 11 k + 5}{(k + 3)!})$ is : [2025]

1 5/3

2 4/3

3 7/3

4 2

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