0
$\begingroup$

For my analysis homework, I am to show that $\lim_{n \to \infty} \frac{3^n}{n!} = 0$ using the epsilon definition. My approach is to invoke the squeeze theorem and show that the above sequence is less than $\frac{3^n}{4^n} \forall \hspace{1mm}n\ge 9$; this I can prove via induction. That is, I want to construct the necessary $N$ from $\left(\frac{3}{4}\right)^n < \epsilon$; and so, $n > \log_{\frac{3}{4}}\epsilon$. My question is, why does the inequality switch after I take the logarithm of both sides? Or does it switch at all?

$\endgroup$
5
$\begingroup$

Since the $\ln$ function is monotonically increasing then

$$\left(\frac34\right)^n<\epsilon\iff n\underbrace{\ln\left(\frac34\right)}_{<0}<\ln(\epsilon)\iff n>\frac{\ln(\epsilon)}{\ln\left(\frac34\right)}=\log_{\frac34}(\epsilon)$$

$\endgroup$
0
$\begingroup$

The inequality "switches" (from $<$ to $>$) because you have to effectively divide by a negative number. This is a law of the inequality sign.

$\endgroup$

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.