Does $n!^{\frac{1}{n}}$ Converge? I am hoplessly trying the ratio test. So if $a_n=n!^{\frac{1}{n}}$ then we compute the limit of the ratio $\frac{a_{n+1}}{a_n}$ if it turns out to be less than 1 it converges. So
$$\frac{(n+1)!^{\frac{1}{n+1}}}{n!^{\frac{1}{n}}}=(n!)^{\frac{1}{n+1}-\frac{1}{n}}(n+1)^{\frac{1}{n+1}}$$.
Now I know that $(n+1)^{\frac{1}{n+1}}$ converges to 1. How do I proceed?
 A: (I added a more general result at the end.)
Here is a completely elementary proof
that does not need
Stirling's formula
or even logs:
$n!
=\prod_{k=1}^n k
>\prod_{k=\lfloor n/2 \rfloor}^n k
>\lfloor n/2 \rfloor^{\lceil n/2 \rceil}
\ge\lfloor n/2 \rfloor^{ n/2}
$.
Therefore,
$n!^{1/n}
>\lfloor n/2 \rfloor^{1/2}
$
which diverges.
Note:
You can be more precise
and get better estimates,
but this is enough
to show divergence.

More generally,
if you take the terms
after $an$,
where $0 < a < 1$,
there are $n(1-a)$ terms,
each at least $an$,
so
$n!
>(an)^{n(1-a)}
$
or
$n!^{1/n}
>(an)^{1-a}
$.
This does not get to
the true value of
$cn$
for some $c$,
but it gets close.

Another tack:
Divide $1$ to $n$
into $k$ parts.
Each part has
$n/k$ numbers
with the smallest
of the $j^{th}$ part being
$nj/k$, 
for $j$ from $0$ to $k-1$.
Therefore,
skipping the first part,
$n!
>\prod_{j=1}^{k-1} (nj/k)^{n/k}
=(n/k)^{n(k-1)/k}\prod_{j=1}^{k-1} j^{n/k}
=(n/k)^{n(k-1)/k}(k-1)!^{n/k}
$
so
$n!^{1/n}
>(n/k)^{(k-1)/k}(k-1)!^{1/k}
$.
For example,
if $k=3$, then
$n!^{1/n}
>2^{1/3}(n/3)^{2/3}
$.
If we take
$k = \sqrt{n}$,
this shows that
$n!^{1/n}
>(\sqrt{n})^{1-1/\sqrt{n}}(\sqrt{n}-1)!^{1/\sqrt{n}}
$,
or
$n!
>(\sqrt{n})^{\sqrt{n}-1}(\sqrt{n}-1)!^{\sqrt{n}}
$.
Don't know if this is any use,
but it's fun.
A: Hint.
You can either use Stirling approximation as suggested. An alternate way if you're not supposed to know Stirling approximation is to go through logarithm:
$$\ln a_n = \frac{1}{n}\sum_{k=1}^n \ln k$$ and then using that logarithm is increasing to compare the sum to an integral.
A: Let's follow what @b00n heT has suggested:
$$n!\sim \sqrt{2\pi n}\left({n\over e}\right)^n$$
So we are left with finding the limit:
$$\lim_{n\to\infty}\left(2\pi n\right)^{1\over 2n}\left({n\over e}\right)$$
Let's look at the two terms of the product
$$\left(2\pi n\right)^{1\over 2n}\to 1$$
$${n\over e}\to \infty$$
So our sequence goes to $+\infty$
