Rearranging for $n$ with a factorial

Question

For my maths course I need to prove that $n!/2^n$ tends to infinity as $n$ tends to infinity. For this I have to rearrange $n!/2^n > ∂$ so that it says $n > ...$.

How to do this? Thanks in advance.

Answer 1

Using $\epsilon-\delta$ to prove this:

$\forall\epsilon>0,\exists\delta$ such that $n>\delta$ implies $\frac{n!}{2^n}>\epsilon$

We choose $\delta=\max(10, \left\lceil\epsilon\right\rceil)$.

See spoilerbox for rest of proof.

Then, $\frac{n!}{2^n}=n\times\frac{(n-1)!}{2^n}>\epsilon\times\frac{(n-1)!}{2^n}=\epsilon\times\frac{1\times2\times3\times4\times5\times6\times7\times\dots\times(n-1)}{2^n}>\epsilon\times\frac{1\times2\times2\times4\times4\times2\times2\times\dots\times2}{2^n}=\epsilon\times\frac{2^{n-4}\times16}{2^n}=\epsilon$

Answer 2

If $n$ is odd: $n=2m+1$, we can write \begin{align*} \frac{n!}{2^n}&=\frac12\biggl(\frac22\frac32\biggr)\biggl(\frac42\frac52\biggr)\biggl(\frac62\frac72\biggr)\dotsm\biggl(\frac{2m}2\frac{2m+1}2\biggr)\\ &\ge\frac12\frac{2^2}{2^2}\frac{4^2}{2^2}\frac{3^2}{2^2}\dotsm\frac{(2m)^2}{2^2}=\frac12 (m!)^2, \end{align*} which tends to $\infty$.

Now, the sequence is clearly increasing, hence if the subsequence of odd terms tends to $\infty$, the sequence itself tends to $\infty$.

Answer 3

$n!$ contains about $n$ factors of $2$, $n/2$ factors of $3$, $n/4$ factors of $5$ and generally about $n/(p-1)$ factors of $p$ for every prime $p$, with an error of $\log_p(n)$.

Thus for instance, $n!/2^n$ grows asymptotically faster than $3^{n/2}$, or $3^{n/2}·5^{n/4}$ etc. Thus it is not surprising that one can find many "bad" estimates that still prove the divergence towards $+∞$.