How to determine the limit of $\{\frac{(n!)^3}{(3n)!}\}$ I just finished an assignment that had me determine if the series
$$
\sum_{n=1}^{\infty}\left(\frac{(n!)^3}{(3n)!}\right)
$$
is convergent or divergent. For this, we have the Ratio Test and everything is great.
But, now, after having done the assignment, I got curious and wanted to try and find the limit of the sequence $\{a_n\}=\{\frac{(n!)^3}{(3n)!}\}$.
I'm not sure how to approach this. Since I came to the result in my assignment that the series above is convergent, the sequence must be null. I have tried solving $|a_n| < \varepsilon$ for $n$, but can't make this work. I also tried using the squeeze rule, but again, not with alot of luck.
How should I approach this?
 A: Same, you can use something similar to the ratio test, but as you said, you could just conclude from the ratio test.
$\exists K\in\mathbb{N}$ s.t. $\forall n\ge K, |{a_{n+1}\over a_n}|={(n+1)!^3\over(3(n+1))!}/{n!^3\over(3n)!}\le{1\over2}$.
$\forall n>K, |a_n|\le{1\over2}|a_{n-1}|\le{1\over2^{n-K}}|a_K| \to 0$ as $n\to\infty$.
A: I'd try splitting up the fraction into three products:
$$\begin{align*}
\frac{(n!)^3}{(3n)!} &= \frac{n!}{(3n)(3n-1)\cdots(2n+1)} \cdot \frac{n!}{(2n)(2n-1)\cdots(n+1)} \cdot \frac{n!}{n!}\\ 
&\le \frac{n!}{(2n)(2(n-1))(2(n-2))\cdots(2\cdot2)(2\cdot1)}\cdot \frac{n!}{n(n-1)(n-2)\cdots(2)(1)} \cdot 1\\
& \le \frac{n!}{2^nn!} \\&
\le \frac{1}{2^n} \to 0 \qquad \mbox{ as } n \to \infty
\end{align*}$$
And also we have that every term in $\left\{\frac{(n!)^3}{(3n!)}\right\}$ is greater than zero, so the sequence converges to $0$ by the squeeze theorem.
A: I assume you mean
$\{a_n\}=\{\frac{(n!)^3}{(3n)!}\}$,
not
$\{a_n\}=\{\frac{(n!)^3}{(3n)}\}$.
More generally,
look at
$\{a_n\}=\{\frac{(n!)^k}{(kn)!}\}$
for an integer $k$.
The key is
Stirling's theorem:
$n!
\approx \sqrt{2\pi n}(n/e)^n
$.
There are more accurate forms,
but this is enough for
this.
$(n!)^k
\approx (\sqrt{2\pi n}(n/e)^n)^k
= (2\pi)^{k/2}n^{kn+k/2}/e^{kn}
$
and
$(kn)!
\approx \sqrt{2\pi kn}(kn/e)^{kn}
=\sqrt{2\pi}k^{kn+1/2}n^{kn}/e^{kn}
$
so
$\frac{(n!)^k}{(kn)!}
\approx\frac{(2\pi)^{k/2}n^{kn+k/2}/e^{kn}}{\sqrt{2\pi}k^{kn+1/2}n^{kn}/e^{kn}}
=\frac{(2\pi)^{k/2-1/2}n^{k/2}}{k^{kn+1/2}}
=\frac1{\sqrt{2\pi k}}\frac{(2\pi)^{k/2}n^{k/2}}{k^{kn}}
=\frac1{\sqrt{2\pi k}}\left(\frac{(2\pi)^{1/2}n^{1/2}}{k^{n}}\right)^k
$
and this tends to zero quite quickly.
