Find $\lim_{n \to \infty} \left( \frac{3^{3n}(n!)^3}{(3n)!}\right)^{1/n}$ Find $$\lim_{n \to \infty} \left( \frac{3^{3n}(n!)^3}{(3n)!}\right)^{1/n}$$
I don't know what method to use, if  we divide numerator and denominator with $3^{3n}$, I don't see that we win something. I can't find two sequences, to use than the squeeze theorem. I'm stuck.
 A: The Stolz-Cesàro Theorem says that if
$$
\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=L
$$
then
$$
\lim_{n\to\infty}\frac{a_n}{b_n}=L
$$
This is the discrete, pre-calculus analog of L'Hôpital.

Suppose we let
$$
a_n=\log\left(\frac{3^{3n}(n!)^3}{(3n)!}\right)
$$
and
$$
b_n=n
$$
Then
$$
a_{n+1}-a_n=\log\left(\frac{3^3(n+1)^3}{(3n+3)(3n+2)(3n+1)}\right)
$$
and
$$
b_{n+1}-b_n=1
$$
From this, we get
$$
\begin{align}
\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}
&=\lim_{n\to\infty}\log\left(\frac{3^3(n+1)^3}{(3n+3)(3n+2)(3n+1)}\right)\\
&=\log\left(\lim_{n\to\infty}\frac{3^3(n+1)^3}{(3n+3)(3n+2)(3n+1)}\right)\\[3pt]
&=\log(1)\\[9pt]
&=0
\end{align}
$$
Therefore,
$$
\begin{align}
\log\left(\lim_{n\to\infty}\left(\frac{3^{3n}(n!)^3}{(3n)!}\right)^{1/n}\right)
&=\lim_{n\to\infty}\frac{\log\left(\frac{3^{3n}(n!)^3}{(3n)!}\right)}n\\
&=\lim_{n\to\infty}\frac{a_n}{b_n}\\
&=\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}\\[3pt]
&=0
\end{align}
$$
From which we get
$$
\lim_{n\to\infty}\left(\frac{3^{3n}(n!)^3}{(3n)!}\right)^{1/n}=1
$$
A: One way is to take logarithms. 
If you take the logarithm, you get $\frac{1}{n} \left( 3n \log 3 + 3 \log n! - \log ((3n)!) \right)$.
Now, Stirling's approximation says $\log n!  = n \log n -n + O(\log n)$. So, $\log ((3n)!) = 3n \log(3n) - 3n + O(\log n)$ and thus $3 \log n! - \log (3n)! = 3 n \log n - 3 n + O(\log n) - \left ( 3n \log(3n) - 3n + O(\log n) \right) = -3n \log 3 + O(\log n)$. Thus, $\frac{ 3 \log n! - \log ((3n)!) } {n} = -3 \log 3 +  \frac{O(\log n)}{n}$. Plugging this into the expression for the logarithm, we have that the logarithm of the desired limit is $\frac{O(\log n)}{n} \to 0$, so the log of the limit is $0$ so the limit is $1$. 
A: Use equivalents  and Stirling's formula: $\;n!\sim_\infty \sqrt{2\pi n}\Bigl(\dfrac n{\mathrm e}\Bigr)^{\!n}$:
$$\biggl(\!\frac{3^{3n}(n!)^3}{(3n)!}\!\biggr)^{1/n}\!\sim_\infty\left(\frac{3^{3n}\sqrt{(2\pi n)^3\strut}\Bigl(\dfrac{n}{\mathrm e}\Bigr)^{\!3n}}{\sqrt{6\pi n}\Bigl(\dfrac{3n}{\mathrm e}\Bigr)^{\!3n}}\right)^{\!\tfrac 1n}=\Bigl(\sqrt{\tfrac 43} \pi n\Bigr)^{\!\tfrac 1n}\to 1.$$
A: Let $a_n = \dfrac{3^{3n}(n!)^3}{(3n)!}$.
The limit $\lim a_n^{1/n}$ is related to the root test for series convergence.
In this context, it is well known that if $\lim \dfrac{a_{n+1}}{a_n}$ exists, then so does $\lim a_n^{1/n}$ and they are equal. Let's compute the ratios then:
$$
\lim \frac{a_{n+1}}{a_n} = \lim 3^3 \dfrac{(n+1)^3}{(3n+3)(3n+2)(3n+1)} = 1
$$
