Find $\lim_{n \to \infty} \sqrt[n]{n!}$ using squeeze theorem.

Question

I am assigned to find $\lim_{n \to \infty} \sqrt[n]{n!}$, but for my course's level of knowledge, we must use the squeeze theorem.

I can easily find the right side of the theorem (a limit that is bigger than the given one) by using the arithmetic mean and geometric mean inequality
$\sqrt[n]{n!}=\sqrt[n]{n(n-1)\dots1}<\frac{n+(n-1)+\dots+1}n=\frac{n(\frac{1+n}2)}n=\frac{1+n}2,$ and we know that $\lim_{n \to \infty} \dfrac{1+n}2=\infty$.

Can you help me find a sequence that is smaller than $\sqrt[n]{n!}$ but has the limit to $\infty$.

Answer 1

Note that

$n!\cdot n! = (1\cdot 2 \cdot \cdots \ \cdot (n-1)\cdot n)(1\cdot 2\cdot \cdots \cdot (n-1)\cdot n)=\left( (1\cdot n)(2\cdot (n-1))\cdots ((n-1)\cdot 2)(n\cdot 1) \right).$

For each $k=1, 2, \cdots, n$,

$k\cdot (n-(k-1))-n=nk-k^2+k-n=n(k-1)-k(k-1)=(n-k)(k-1)\geq 0$

which implies that $k\cdot (n-(k-1))\geq n$.

Thus,

$(n!)^2\geq n^n \implies (n!)^{1/n}\geq \sqrt{n}$

Does this help?

Answer 2

Hint: You can use Stirling's approximation:

$ \sqrt{2 \pi n}\ \left(\frac{n}{e}\right)^n e^{\frac{1}{12n + 1}} < n! < \sqrt{2 \pi n}\ \left(\frac{n}{e}\right)^n e^{\frac{1}{12n}}. $