Question

determine the limit of the sequence or show that the sequence diverges. if it converges, find its limit.

\\a_n = \frac{(1,000)^n}{n!}\\

\\\frac{a_{n+1}}{a_n} = \text{box} \to \text{box} \text{ as } n \to \infty. \text{ so } a_n \to \text{box}. \text{ therefore, the sequence } \text{select}.\\

Explanation:

Compute the ratio of consecutive terms

$$ \frac{a_{n+1}}{a_n} = \frac{\frac{(1,000)^{n+1}}{(n+1)!}}{\frac{(1,000)^n}{n!}} = \frac{(1,000)^{n+1}}{(1,000)^n} \cdot \frac{n!}{(n+1)!} = \frac{1,000}{n+1} $$

Evaluate the limit of the ratio

$$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{1,000}{n+1} = 0 $$

Determine sequence convergence

$$
LATEXBLOCK0
$$

Answer:

Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.

$$a_n = \frac{(1,000)^n}{n!}$$

\(\frac{a_{n+1}}{a_n} =\) <blank>\(\frac{1,000}{n+1}\)</blank> \(\to\) <blank>\(0\)</blank> as \(n \to \infty\). So \(a_n \to\) <blank>\(0\)</blank>. Therefore, the sequence <blank>converges</blank>.