# Proof that limit as n approaches inf of $3^n \cdot \frac{1}{n!} = 0$ using squeeze theorem

This is a homework question. I'm completely new to epsilon proofs, so I'm pretty bad at them.

I want to prove that $3^n \cdot \frac{1}{(n!)} = 0$ converges to 0.

Here's where I'm at. By the definition of a limit of a sequence, I want

$\forall \epsilon>0 \exists N \in \mathbb{N}$ such that $\mid \frac{3^n}{n!} - 0 \mid < \epsilon$.

Or, I want to use the squeeze theorem to bound it below and above with functions with limits of 0.

I started to try to get an appropriate value for N, but it doesn't seem to work out... I'm using the arbitrarily small $\epsilon$ to try to find an N for which $\frac{3^n}{n!} < \epsilon$ for all n > N. I've just tried doing algebraic manipulation, but I haven't got anywhere.

For the squeeze theorem, I thought of trying to bound the sequence between the sequences $x_n = \frac{1}{n}$ and $z_n = 0$. But I would have to prove that $\frac{3^n}{n!}$ is bounded above by $\frac{1}{n}$, which I'm not sure I know how to do.

If you could tell me which of the two methods would be appropriate and some hints for how to proceed I would appreciate it, thanks.

$$0 \leq \frac{3^n}{n!} = \frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot ... \cdot \frac{3}{n} \leq \frac{3}{1} \cdot \frac{3}{2} \cdot \frac{3}{3} \cdot \frac{3}{n}=\frac{27}{2n}$$

• I think it should be $\frac{27}{2n}$, but yes, elegant answer indeed! Thank you. – user197696 Jan 23 '15 at 16:58
• @user197696 ty, fixed it :) – N. S. Jan 23 '15 at 17:07

for large $n$ ;

$\dfrac{1}{n!}< \dfrac{3^n}{n!}< \dfrac{1}{2^n}$

Pick an $\epsilon>0$, can we find an $n:3^n<\epsilon n!$?

Take log of both sides:

$$n\ln(3)<\ln(\epsilon)+\ln(n!)=\ln(\epsilon)+\sum\limits_{i=1}^n \ln(i)<\ln(\epsilon)+n(n+1)/2$$

Therefore,

$$n\ln(3)<\ln(\epsilon)+n(n+1)/2 \implies n(2\ln(3)-1)-n^2>2\ln(\epsilon)$$

Thus, $\exists c: \frac{3^c}{c!}<\epsilon, \; \forall n>c$

• How do you know that $\sum\limits_{i=1}^n \ln(i)<n$? Isn't that false for n=15, for example? ln(15!) = 27.8993, which is not less than 15... – user197696 Jan 23 '15 at 5:36
• @user197696 yes, it was a mistake. I corrected to bound by the first $n$ integers, not $n$. – user76844 Jan 23 '15 at 5:56