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How do I prove convergence of below alternating sequence with a factorial in the denominator?

$\{\frac{\left(-5\right)^n}{n!}\}$

How can I apply the principles described in this post to this question? What sequences can I use to apply Squeeze Theorem?

cashsky
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  • By using that post you can show separately that $-5^n/n! \to 0$ and $5^n/n! \to 0$; the result follows by the Squeeze Theorem. – Connor Harris Apr 09 '18 at 01:26
  • Once you have $5^n/n! \to 0$, $(-5)^n/n! = (-1)^n5^n/n!$ also $\to 0$. – marty cohen Apr 09 '18 at 02:20
  • I don't agree that the given is a duplicate of the linked OP since here the key point asked is how to apply the squeeze theorem. I'm asking an advice by moderators about that. – user Apr 09 '18 at 12:08

1 Answers1

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Note that

$$\frac{\left(-5\right)^n}{n!}=(-1)^n\frac{5^n}{n!}\to0$$

indeed

$$-\frac{5^n}{n!}\le (-1)^n\frac{5^n}{n!}\le \frac{5^n}{n!}$$

and by ratio test

$$\frac{5^{n+1}}{(n+1)!}\frac{n!}{5^n}=\frac{5}{n+1}\to 0$$

user
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