I found on some online sites that
$$ \lim_{ n \to \infty} \sqrt[n]{n!}= \infty $$
But I did not undertand ow they get it.
My attempt: $$ \lim_{ n \to \infty} \sqrt[n]{n!}= \lim_{ n \to \infty} e^{\ln[ (n!)^{\frac1n}]} = \lim_{ n \to \infty} e^{\frac1n \ln(n!)} = \lim_{ n \to \infty} e^{ \frac{\ln(n)}{n} +\frac{\ln(n-1)}{n}+ \cdots } = e^{ 0 +0+ \cdots }= e^0 =1 $$
On
As mentioned above, each term of the sequence $$ \frac{\ln n}{n} + \frac{\ln(n-1)}{n} + \cdots + \frac{\ln 1}{n} $$ tends to zero, but you have $n$ of them. What is the sum of many small things? Well, it depends how many, and how small.
To solve this, you can indeed use Stirling's formula. Without it, a simple way is to see that, for $m_n = \lfloor n/2 \rfloor$ $$ n! = n \times (n-1) \times \cdots \times 1 \geq n \times (n-1) \times \cdots \times m_n \geq m_n^{m_n}, $$ and thus $$ \frac{\ln (n!)}{n} \geq \frac{\ln(m_n^{m_n})}{n} = \frac{m_n}{n} \ln m_n \to + \infty, $$ since $m_n/n \to 2$ and $\ln m_n \to + \infty$.
$n! > (n/2)^{(n/2) };$
$(n!)^{(1/n)} >(n/2)^{(1/2)};$
Hence?