I would like to prove that the sequence $ {a_n} = \frac{n!}{2^n}$ diverges to $+ \infty$. As I understand it, this means that for all numbers $M$, I must find a number $N$ such that for all $n \ge N$, I get $a_n \ge M$. However, I'm not sure how to pick $N$.
Thanks.
Answer
You can use a lower bound on $n!$ instead of $n!$ itself. For example, when $n \ge 3$, $a_n = \frac{n!}{2^n} = \frac{2}{2^n}(3\cdot \ldots \cdot n) \ge \frac{2}{2^n}(3^{n-2}) = \frac{1}{2}\left(\frac{3}{2}\right)^{n-2}$. Since the sequence $\left(\frac{3}{2}\right)^n$ goes to $+\infty$ as $n$ goes to $+\infty$, $a_n$ must also go to $+\infty$.
If you wish to go all the way back to the definition, you only need to elaborate the last sentence. This can be done as follows.
Note that now we know that $a_n \ge \frac 12\left(\frac 32\right)^{n-2}
$ for $n \ge 3$. For any given $M > 0$, pick $N$ such that $\frac 12\left(\frac 32\right)^{N-2} = M$. (More specifically, $N = \frac{\log(2M)}{\log(3/2)}+2$.) If $N < 3$, we pick $N = 3$ instead. Then, for all $n > N$, we know $a_n \ge \frac 12\left(\frac 32\right)^n > M$.
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