Having a hard time with Calculus 1. I actually think I do understand the concept of limit, but I get stuck too much often when exercising. Perhaps mine is more of an algebra knowledge problem? Anyway I'd like to know how to prove the limit of this sequence. $$\lim_{n\to +\infty} \frac{n!}{q^n} = +\infty \ \forall q>1$$
My first thought was to start proving that at some point $\frac{(n+1)!}{q^{n+1}}>\frac{n!}{q^n}$ but I don't really know how could I do that.
Suppose $N\leq q\leq N+1$ so define $a_n=\frac{n!}{q^n}$
and now rewrite $$a_n=\frac{1\times2\times3\cdots \times N\times(N+1)\cdots \times n}{q\times q \cdots \times q}$$ now look again for $n\geq N+1$ can you take over?
$$ a_n=\underbrace{\frac 1q}_{\leq1} \times \underbrace{\frac 2q}_{\leq1}\times \cdots \underbrace{\frac Nq}_{\leq1}\times \underbrace{\frac {N+1}q}_{\geq 1}\times \cdots\times\underbrace{\frac nq}_{\geq1}$$
For $ q \in \mathbb{R} $ and $ q > 0 $,
by Archimedian principle, $ \; \exists \; k \in \mathbb{N} $ such that $ k > q $
Then, for $ n > k $, $$ q \cdot q \cdot q \dots q < k \cdot (k+1) \cdot (k+2) \dots (n-1) \\ \implies q^{n-k} < \frac{(n-1)!}{(k-1)!} \\ \implies \frac{n!}{q^n} > \frac{n!}{q^k} \cdot \frac{(k-1)!}{(n-1)!} = \frac{(k-1)!}{q^k} \cdot n $$
Now, for any $ M \in \mathbb{R} $,
by Archimedian principle, $$ \exists \; n_0 \in \mathbb{N} \; \text{such that} \; n_0 > M \cdot \frac{q^k}{(k-1)!} $$
Then, $$ \forall n > n_0, \; \frac{n!}{q^n} > \frac{(k-1)!}{q^k} \cdot n > \frac{(k-1)!}{q^k} \cdot n_0 > M $$
Therefore, by definition of limit, $$ \lim_{n \to \infty} \frac{n!}{q^n} = \infty $$
Moreover, this is not only true for $ q > 0 $, but for any $ q \in \mathbb{R} $ which can be proven in a similar manner. Just note that $ |q| > q $, so $ \frac{n!}{q^n} > \frac{n!}{|q|^n} $ and then proceed with the above method.
