Proving the limit of $\frac{n!}{10^{n}}$ using definitions

Question

$\cdot \lim \limits_{n \to \infty} \frac{n!}{10^n} = \frac{10!}{10^{10}} * (\frac{n!}{10^n})$ for all $n \ge 11$

So we must find $N(M)$ such that $\lvert \frac{10!}{10^{10}} * (\frac{n!}{10^n}) \rvert = \frac{10!}{10^{10}} * (\frac{n}{10})< M$, $\forall n>N$

So $\frac{10!}{10^{10}} < \frac{10}{n} * M (\forall n\ge 11)$

Therefore $\frac{1}{M} *\frac{9!}{10^{10}}< n $

And my function can be max{$11,\frac{1}{M} *\frac{9!}{10^{10}} $}?

Answer 1

I don't understand why you write $$ \lim_n\frac{n!}{10^n}=\frac{10!}{10^{10}}\times\frac{n!}{10^n},\quad \forall n\geq 11. $$ A simple and related approach to this problem is as follows: define $\delta=\frac{10!}{10^{10}}>0$. Note also that for $n\geq 11$, $$ \frac{11\times 12\times \cdots \times n}{10^{n-10}}\geq 1.1^{n-10}. $$ This suggests that for any given $M>0$, if we let $N=\max\{11,10+\frac{\log(M/\delta)}{\log 1.1}\}$, then for $n\geq N$, $$ \frac{n!}{10!}=\delta\times\frac{11\times 12\times \cdots \times n}{10^{n-10}} \geq\delta (1.1)^{n-10}>\delta(1.1)^{\frac{\log(M/\delta)}{\log 1.1}}=M. $$

---

Alternative: not using $\log$. Note that: for $n\geq 11$, the Bernoulli's inequality gives $$ 1.1^{n-10}=(1+0.1)^{n-10}\geq 1+\frac{n-10}{10}>\frac{n-10}{10}. $$ Now pick $N=\max\{11,10+\frac{10M}{\delta}\}$.