0
$\begingroup$

By first showing that

$\frac{2^n}{n!}<18(\frac{2}{3})^n \quad$

for all n is an element of real numbers, find

$\lim \limits_{n \to \infty} \frac{2^n}{n!}$

by squeeze theorem.

I'm already stuck trying to prove the first part...I managed to resolve the inequality to:

$\frac{2^n}{n!}<\frac{2^{n+1}}{3^{n-2}}$

$\endgroup$
  • $\begingroup$ Lim x -> 1? Where does x come into it and the squeeze theorem usually takes the limit to infinity. $\endgroup$ – Phil H Jun 3 '18 at 18:05
  • $\begingroup$ 1) I assume that you mean "n is an element of the natural numbers" instead of the real numbers 2) The way the first inequality is given is already a 'good' form, your simple rearrangment is more hiding the way to prove it than revealing something new! Hint: try mathematical induction. $\endgroup$ – Ingix Jun 3 '18 at 18:09
-1
$\begingroup$

$18(\frac{2}{3})^n = 18(\frac{2^n}{3^n})$ which, when compared to $(\frac{2^n}{n!})$ is always bigger. $3^n$ is smaller than $n!$ for $n>6$ and hence $(\frac{2^n}{3^n})>(\frac{2^n}{n!})$ here. Below that, the $18$ factor ensures $18(\frac{2^n}{3^n})>(\frac{2^n}{n!})$

$\endgroup$
  • $\begingroup$ Sorry for the typos. $\endgroup$ – Phil H Jun 3 '18 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.