real analysis - Find the limit of the sequence $frac{n!}{2^n}$ as n tends to infinity

We've only been taught to find limits using the Squeeze Theorem and L'Hopitals Rule, so I'm not sure how to go about finding the limit of this sequence.

Answer

For a simpler idea, note that

$$\frac{a_{n+1}}{a_n}=\frac{n+1}2$$

It follows that $a_{n+1}/a_n>2$ for $n=3,4,\dots$ so that $a_n$ grows faster than $2^n$; thus it diverges to $+\infty$.

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Hint Set

$$a_n=\frac{n!}{ 2^n}$$ and show that $a_n$ is strictly increasing for $n\geq 3$. That is, look at when $${a_{n+1}}>{a_n}$$ is true.

$2^n$ is tiny compared to $n!$ since we're multiplying by a constant factor of $2$, while in $n!$ we're constantly increasing the factor by $1$.

LEMMA

$$n!/2^{n-1}$$ is a (positive) integer for infinitely many values of $n$.

P Take $n=2^k$ for $k=0,1,2,\dots$. The multiplicity for which $2$ divides $2^k!$ is

$$v_2(2^k)=\sum_{m=1}^\infty\left\lfloor \frac{2^k}{2^m}\right\rfloor\\=\sum_{m=1}^\infty\lfloor 2^{k-m}\rfloor\\=\sum_{m=1}^k 2^{k-m}=1+2+\cdots+2^{k-1}=2^k-1$$

which means $2^{2^k-1}\mid 2^k!$.

SPOILER

You should be able to show that

$$a_n\to +\infty$$