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[tex]\lim_{n \to \infty} \frac{n!}{a^{2n+1}} = \infty \qquad a \in \mathbb{N}[/tex]

Here is my try:

Take [tex] a_n = a^{2n+1}/n![/tex] . Since

[tex]

\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{a^{2(n+1)+1}/\,(n+1)!}{a^{2n+1}/\,n!} = \lim_{n \to \infty} \frac{a^{2}}{n+1} = 0 < 1 [/tex](D'Alembert test),

the series [tex]\sum_{n=0}^\infty a_n[/tex] is convergent, i.e.[tex]

\lim_{n \to \infty} a_n = 0.

[/tex]

Then

[tex]

\lim_{n \to \infty} \frac{n!}{a^{2n+1}} = \lim_{n \to \infty} \frac{1}{a_n} = \infty.

[/tex]

Correct? Is there another way to prove it?

Thanks.

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# Homework Help: Attempt at solving a limit

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