How do I solve $\lim_{n \to \infty} \frac{n!}{2^{n+1}}$

Question

I am trying to solve the limit $\lim_{n \to \infty} \frac{n!}{2^{n + 1}}$.

Since both $n!$ and $2^{n + 1}$ approach $\infty$ as $n \to \infty$, my thinking was that I had to take the derivative of both the numerator and the denominator. However, from my research, the derivative of $n!$ is very complicated, involving some gamma function that we haven't covered in the course yet.

Is there an easier, more intuitive approach to solving this problem without using L'Hopital's Rule?

Answer 1

$n! >(1)(2)(3^{n-2})$. Can you finish?

Answer 2

Hint: use the Stirling's approximation, which says that $n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$

Answer 3

This sequence diverges and there's no need for any fancy techniques.

I will prove a more general statement: Let $(x_n)_{n\in\mathbb{N}}$ be a sequence with $x_n=\frac{a^{n+1}}{n!}$ with $a>0,\ a\in\mathbb{R}$.

Now let us take the natural number $N\geq2|a|$. For any $n\geq N$ the following is true: \begin{equation*} \frac{a^{n+2}}{(n+1)!}=\frac{a^{n+1}}{n!}\cdot \frac{a}{n+1}\leq \frac{1}{2}\frac{a^{n+1}}{n!} \end{equation*} Throug induction we can prove that $\frac{a^{n+1}}{n!}\leq 2^{-(n-N)}\frac{a^{N+1}}{N!}$. The right side converges to $0$. Therefore $x_n\to 0$. This yields that your sequence diverges.