$\lim_{n\to\infty} \frac{3^n}{(1+3)(1+3^2)(1+3^3)\ldots (1+3^n)}$ It is visible that the result is 0, but I can't calculate it.
$\lim_{n\to\infty} \frac{3^n}{(1+3)(1+3^2)(1+3^3)\ldots (1+3^n)}$
It occurred to me to express it as a product and take a ratio test, but I'm not sure
 A: Note that $$\frac{3^n}{(1+3)(1+3^2)\dots(1+3^n)}\leq\frac{3^n}{(1+3^{n-1})(1+3^n)}$$
and $$\lim_{n\to\infty}\frac{3^n}{(1+3^{n-1})(1+3^n)}=\lim_{n\to\infty}\frac{1}{1+3^{n-1}}\lim_{n\to\infty}\frac{3^n}{1+3^n}=0\cdot 1=0$$
Hence by the squeeze theorem our original limit is $0$.
A: Hint
$$\begin{align}0≤\lim_{n\to\infty} \frac{3^n}{(1+3)(1+3^2)(1+3^3)\ldots (1+3^n)}≤\lim_{n\to\infty} \frac{3^n}{3×3^2\ldots 3^n}=\lim_{n\to\infty}\dfrac {3^n}{3^{\frac {n(n+1)}{2}}}=0\end{align}$$
A: $$
f(n)=\frac{3^n}{(1+3)(1+3^2)\cdots(1+3^n)}
=\frac{\frac{3^n}{3^1 3^2 \cdots 3^n}}{\frac{1+3^1}{3^1}\frac{1+3^2}{3^2}\cdots\frac{1+3^n}{3^n}}\\
=\frac{\frac{3^n}{3^{\frac{n(n+1)}2}}}{(1+3^{-1})(1+3^{-2})\cdots(1+3^{-n})}
=\frac{2\cdot 3^{-\frac{n(n-1)}{2}}}{\left(-1;\frac{1}{3}\right)_{n+1}}
$$
in terms of a $q$-Pochhammer symbol in the denominator.
For large $n$ this denominator approaches a constant,
$$
\lim_{n\to\infty}\left(-1;\frac{1}{3}\right)_{n+1}
= \left(-1;\frac{1}{3}\right)_{\infty}
\approx 3.129868037134023\ldots,
$$
in Mathematica,
    QPochhammer[-1, 1/3] // N

and therefore $\lim_{n\to\infty}f(n)=0$.
