Show that $6^n/n! \le 6^5/5! \times 6/n$

Question

I want to show that

$$\frac{6^n}{n!} \le \frac{6^5}{5!} \cdot \frac 6n$$

without using induction, which I've done but is rather clunky. Is there a more straight forward way of doing this?

Answer 1

For $n>5$ $$\frac{6^n}{n!} = \frac{6^6 6^{n-6}}{5!6\cdots n}= \frac{6^6}{5!}\frac66\frac67\cdots\frac{6}{n-1}\frac1n \le\frac{6^5}{5!}\cdot\frac 6n,$$ and you can check the cases $n=1,\dots,5$.

Answer 2

We know that $(n-1)(n-2)\cdots 6\geq 6\cdot 6\cdots 6$ with $n\geq 7$ (leaving only finitely many cases to check), where the number of factors on the left are the same as on the right (exactly $n-6$). So $$\frac{(n-1)!}{5!}\geq 6^{n-6}$$ Rewriting this, using $6^{n-6}=\frac{6^n}{6\cdot 6^5}$ and $n!=n\cdot (n-1)!$, gives $$\frac{6^n}{n!}\leq \frac{6^5}{5!}\frac{6}{n}$$

Answer 3

For $n\geq 6\cdots (1)$ $$6^n\geq6^5\cdots (2)$$ $$n!\geq5!\cdots (3)$$ Now simply combine the 3 equations

$$\frac{6^n}{ 6^5}\frac{n!}{5!} \frac{n}{6}\geq 1$$

Answer 4

$$\frac{6^n}{n!} = \prod_{i=1}^n \frac{6}{i}$$ Is increasing for $n<6$, constant for $n=6$ and decreasing $n>6$

Hence $$\frac{6^6}{6!} = \frac{6^5}{5!}$$ is a maximum.