limit of sequence $\{a_n\}$ where $a_n = n!/c^n$, as $n$ tends to infinity

Question

I was wondering if there was a general way to find (or show that the sequence diverges) $$ \lim_{n \to \infty} \{a_n\}$$ Where $a_n$ is of the form $\frac{n!}{c^n}$, $c$ being a non-zero natural number.

I think that I was able to find an intuitive answer following the reasoning below. However, I would like to know if there is a more rigorous way to go about it.

We can show that past a certain point, the sequence is growing if $\frac{a_{n+1}}{a_n} \geq 1$ for all $n$ greater than some value: $$ \begin{align} \frac{a_{n+1}}{a_n} &\geq 1 \\\\ \frac{(n+1)!}{c^{(n+1)}}\cdot\frac{c^n}{n!} &\geq 1 \\\\ \frac{n+1}{c} &\geq 1 \\\\ n &\geq c-1 \end{align} $$ So, after $c-1$ terms, the sequence will start growing.

Also, because of the definition of factorials, at the cth term, both the numerator and denominator are going to be multiplied by $c$. The next term will then be multiplied by $\frac{c+1}{c}$, etc. In other words, after $c$ terms, the numerator will start growing faster than the denominator. Because of this, the sequence will grow faster and faster.

Considering this, my assumption is that the sequence cannot have an upper bound, and because it is growing, it must diverge to infinity.

Bonus question: What if $c$ was any real number other than zero? I'm guessing that a negative value would cause oscillation between $-\infty$ and $\infty$.

Answer 1

I’d say that the approach you are taking is correct. One way to maybe make it more clear could be the following.

We’d like to prove that $a_n\to \infty$. Then, for $n \geq c$, we have $$\frac{n!}{c^n} = \frac{c!}{c^c} \frac{(c+1)\cdots n}{c^{n-c}} \geq \frac{c!}{c^c} \left( \frac{n+1}{c} \right)^{n-c}$$ Where on the right hand side, the first term is a constant, and the second is the power of a number greater than one. Therefore, as $n \to \infty$, the sequence must also tend to infinity.

Your answer for the bonus question is correct.