I am struggling to find $$\lim_{n\rightarrow\infty}n!^{1/n}$$
being trying this for ages, I know that:
$$\lim_{n\rightarrow\infty} n^{{1}/{n}} = 1$$
which might be useful for using pinching theorem. but I can't think of a way to 'trap' the function underneath.
Let
$$ a_n = \left(n!\right)^{1/n} . $$
Then
$$ \begin{align}\ln a_n &= \frac 1n \ln n! \\ \ln a_n &= \frac 1n \ln \left(\prod_{k=1}^n k\right) \\ \ln a_n &= \frac 1n \sum_{k=1}^n \ln k \\ \end{align} $$ or, since $\ln 1=0$,
$$ \ln a_n = \frac 1n \sum_{k=2}^n \ln k . $$
The sum can be viewed as an upper Riemann sum. (We could also make it a lower Riemann sum, but it will turn out we don't need to.) Since $\ln t < \ln k$ when $t \in [k-1,k)$,
$$ \begin{align} \int_1^n \ln t \, dt &< \sum_{k=2}^n \ln k \\ \sum_{k=2}^n \ln k &> \big[ t \ln t - t \big]_{t=1}^{t=n} \\ \sum_{k=2}^n \ln k &> n \ln n - n + 1 \\ \frac 1n \sum_{k=2}^n \ln k &> \ln n - 1 + \frac 1n \\ \ln a_n &> \ln n - 1 + \frac 1n \end{align} $$
So $\ln a_n$ and therefore $a_n$ are unbounded increasing sequences.
$$ \lim_{n \rightarrow \infty} \left(n!\right)^{1/n} = \infty $$
