Show that $\frac{3^n}{n!}$ converges to $0$

Question

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!

Answer 1

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.

Answer 2

Hint:
You can use the squeezing principle:

Set $u_n=\dfrac{3^n}{n!}$. Show that $\;\dfrac{u_{n+1}}{u_n}\le\dfrac34$ for all $n\ge 3$. Deduce that $$u_n\le \Bigl(\frac34\Bigr)^{n-3}u_3\enspace\text{ for all }\;n\ge 3.$$

Answer 3

Here is an elementary proof that does not require bringing the log function into the picture.

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Set $u_n=\dfrac{3^n}{n!}$. Observe that $u_0 = 1$, $u_1 = 3$ and $u_2 = 4.5$. We claim that for every $n \ge 0$,

$\tag 1 |u_n| \le 4.5$

If not then let $k$ be the smallest integer such that $|u_k| \gt 4.5$. Now $k$ must be greater than $2$ and we can write $u_k = u_{k-1} \, \dfrac{3}{k}$ so that $u_k \le u_{k-1}$. But this gives a contradiction since $u_{k-1} \le 4.5$.

So $\text{(1)}$ is always true. Let $\varepsilon \gt 0$ be given and let $n \gt 0$ so that

$n \gt \frac{13.5}{\varepsilon} \text{ implies }$

$(4.5) \, \frac{3}{n} \lt \varepsilon \text{ implies }$

$ u_{n-1} \, \frac{3}{n} \lt \varepsilon \text{ implies }$

$ u_n \lt \varepsilon$

Answer 4

Here is another possible view. You have $$ \frac{3^n}{n!}=\frac31\,\frac32\,\overbrace{\frac33\frac34\cdots\frac3{n-1}}^{\text{each of these is at most 1}}\,\frac3n\leq \frac92\times1\times\frac3n=\frac{27}{2n}\leq\frac{14}n. $$ So if $n>14/\varepsilon$, we obtain $3^n/n!<\varepsilon$.

This achieves an easier estimate, at the cost of hiding the fact that the sequence converges a lot faster.