$\lim_{n\to\infty} \frac{3^n}{(1+3)(1+3^2)(1+3^3)\ldots (1+3^n)}$ 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

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 $0$.
