little O proof wrong approach

I am trying to show that $3^n \in o(n!)$ via limits i.e. I want to show that $\lim_{n-> +\infty} \frac{ 3^n}{n!}=0$

To find the $\lim_{n-> +\infty} \frac{ 3^n}{n!}$ I write $\frac{ 3^n}{n!}= e^{\ln({\frac{3^n}{n!}})}= e^{\ln(3^n)-ln(n!)}=e^{n\ln 3- \ln n-\ln (n-1)...-\ln2}= e^{n [\ln 3- \frac{\ln n}{n}-\frac{\ln (n-1)}{n}-...-\frac{\ln2}{n}]}$ and hence $\lim_{n-> +\infty}e^{n [\ln 3- \frac{\ln n}{n}-\frac{\ln (n-1)}{n}-...-\frac{\ln2}{n}]}$= $e^{+\infty (\ln 3)}=+\infty $, which is obviously not zero. (I am using that as $\lim_{n-> +\infty}\frac{ln(n)}{n}=...=\lim_{n-> +\infty}\frac{ln(2)}{n}=0$)

Where am I going wrong in my approach? Alternatively if someone can provide me with a hint how to show $3^n \in o(n!)$ via limits. Thank you in advance.

The problem is that when you replace $\ln 3 - \frac{\ln n}{n} - \frac{\ln (n-1)}{n} - \dots - \frac{\ln 2}{n}$ by a single $\ln 3$, you are getting rid of $n-1$ terms. Individually, those terms may go to $0$, so we may drop one, or two, or any constant number of them in the limit. But the number of these terms goes to $\infty$, so it's not safe to drop them.

To deal with the $n!$, you might use the squeeze theorem: this requires an upper bound on $\frac{3^n}{n!}$ (which still goes to $0$), so it requires a lower bound on $n!$. I suggest proving one of the following, your choice:

• $n! \ge 4^{n-3}$, in which case $\lim_{n \to \infty} \frac{3^n}{n!} \le \lim_{n \to \infty} \frac{3^n}{4^{n-3}} = \lim_{n \to \infty} 64 (\frac34)^n = 0$.
• $n! \ge (\frac n2)^{n/2}$, in which case $\lim_{n \to \infty} \frac{3^n}{n!} \le \lim_{n \to \infty} \frac{3^n}{(\frac n2)^{n/2}} = \lim_{n \to \infty} (\frac{18}{n})^{n/2} = 0$.

Stronger bounds exist. For example, we can prove that $n! \ge (\frac ne)^n$ for all $n$, which gets quite close to Stirling's formula: $n! \sim \sqrt{2\pi n} (\frac ne)^n$. Either of these would also be good enough here, but they're overkill.