Rearranging for $n$ with a factorial 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. 
 A: 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  $\infty$.
A: 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, $\frac{n!}{2^n}=n\times\frac{(n-1)!}{2^n}>\epsilon\times\frac{(n-1)!}{2^n}=\epsilon\times\frac{1\times2\times3\times4\times5\times6\times7\times\dots\times(n-1)}{2^n}>\epsilon\times\frac{1\times2\times2\times4\times4\times2\times2\times\dots\times2}{2^n}=\epsilon\times\frac{2^{n-4}\times16}{2^n}=\epsilon$

A: $n!$ 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 $+∞$.
