Prove limit is zero $A_n = \frac{(1/3)(1/3+1)\dots(1/3 + (n-1))}{n!}$

I came across this sequence $A_n$ while studying the convergence of the Maclaurin series of this function

$$f(x) = \frac{x}{\sqrt[3]{1+x^2}}$$

When I write down the Maclaurin series, it turns out it's absolutely convergent for $|x|\lt1$

OK... But what happens at $x=\pm1$? I think at these two points the Maclaurin series turns into an alternating number series. To prove these two series (for $x=\pm1)$ are convergent, I need to show that the following sequence goes to zero and use the Leibniz criterion.

$$A_n = \frac{(1/3)(1/3+1)\dots(1/3 + (n-1))}{n!}$$

But I think I have no idea how to do it.

I see the sequence decreases but how do I show it goes to zero?
I think there should be simple trick to prove that the limit of this sequence is zero.

Maybe I need to show $A_n < 1/\ln(n)$ or something like that, not sure?!

One has $$\log(A_n) = \sum_{k=0}^{n-1} \log \left( \frac{\frac{1}{3} + k}{k+1}\right) = \sum_{k=0}^{n-1} \log \left( \frac{3k+1}{3k+3}\right)= \sum_{k=0}^{n-1} \log \left( 1-\frac{2}{3k+3}\right) $$

Now, $$\log\left( 1-\frac{2}{3k+3}\right) \sim - \frac{2}{3k+3}$$

which is the general term of a divergent series. Hence, by comparison, the series $\displaystyle{\sum \log \left( 1-\frac{2}{3k+3}\right) }$ is also divergent, and because it has negative terms, you get that $$\sum_{k=0}^{+\infty} \log \left( 1-\frac{2}{3k+3}\right) = - \infty$$

or equivalently, $$\lim_{n \rightarrow +\infty} \log(A_n) = -\infty$$

which leads directly to $$\boxed{\lim_{n \rightarrow +\infty} A_n = 0}$$

I was able to prove by induction that

$$A_n \lt \frac{1}{\ln{n}} \tag{***}$$

for all $n \ge 6$

The tricky part is to start the induction at $n=6$ (the base case of the induction) so that the induction step can be done smoothly later on.

The rest of the induction follows in a pretty straightforward way. We assume that the inequality is true for some $n \ge 6$, and we prove it for $n+1$ using the assumption we have for $n$.

The only thing I use is that $$\left(1+\frac{1}{n}\right)^n \lt 3$$ for all $n \in \mathbb{N}$

In fact $(***)$ is also true for $n = 2,3,4,5$ but it seems one has to check these cases separately one by one.

From that inequality $(***)$ and since $A_n \gt 0$, it follows easily that $${\lim_{n \rightarrow +\infty} A_n = 0}$$