<s>[INST] Show that $\frac{3^n}{n!}$ converges to $0$ I was wondering if this proof is correct.
$$\left|\frac{3^n}{n!} - 0\right| = \frac{3^n}{n!} \lt \frac{3^n}{2^n} \lt \varepsilon$$
So then
$$\frac{2^n}{3^n} \gt \frac{1}{\varepsilon}$$
$$\left(\frac{2}{3}\right)^n \gt \frac{1}{\varepsilon}$$
$$\log\left(\frac{2^n}{3^n}\right) \gt \log\left(\frac{1}{\varepsilon}\right)$$
$$n\log\left(\frac{2}{3}\right) \gt \log\left(\frac{1}{\varepsilon}\right)$$
$$n \gt \log\left(\frac{1}{\varepsilon}\right)/\log\left(\frac23\right)$$
So then any $N \gt \log(\frac1\varepsilon)\log(\frac23)$ yields the result we want, so that for all $n \gt N, \left|\frac{3^n}{n!} - 0\right| \lt ε$
Thanks in advance!
 [/INST] You can modify your argument as follows:
You can assume that $n \geq 4$ because shifting the terms of a sequence does not affect its convergence or the value of convergence.
Then 
$$\left|\frac{3^n}{n!} - 0\right| = \frac{3^n}{n!} =\frac{27}{3\times 2\times 1} 
\frac{3^{n-3}}{n \times(n-1)\times\cdots\times 4}\lt 5 \times \frac{3^{n-3}}{4^{n-3}} \lt \varepsilon$$
Now rewrite your own argument to finish the proof:
$$(\frac{3}{4})^{n-3} < \frac{\varepsilon}{5}$$
$$\ln(\frac{3}{4})^{n-3} < \ln\frac{\varepsilon}{5}$$
$$(n-3)\ln(\frac{3}{4}) < \ln\frac{\varepsilon}{5}$$
$$n \geq \lceil \ln\left(\frac{\varepsilon}{5}\right)/\ln\left(\frac{3}{4}\right) \rceil + 3$$
where $\lceil \cdot \rceil$ denotes the ceiling function.
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