How to prove the 2nd question of sequence and series

Question

$$a_1 = \frac32\quad\text{and}\quad\forall n \ge 2,\quad a_n = \frac{3na_{n-1}}{2a_{n - 1} +n - 1}.$$

1. Get $a_n$
2. Prove: $$\forall n \in\Bbb N,\quad a_1a_2\dots a_n < 2\cdot n!$$

Here is what I have gone so far:

$$a_n = \frac{n3^n}{3^n - 1}$$

But I do not know how to prove the question 2.

I found that it is equivalent to proving:

$$ \left(1-\frac13\right)\left(1-\frac1{3^2}\right)\dots\left(1-\frac1{3^n}\right)> \frac12. $$

Answer 1

Using $(1 - x)(1 - y) = 1 - x - y + xy > 1 - x - y$ for all $x, y > 0$, we have \begin{align*} &\left(1-\frac13\right)\left(1-\frac1{3^2}\right)\dots\left(1-\frac1{3^n}\right)\\[6pt] >{}& \left(1 - \frac13 - \frac{1}{3^2}\right)\left(1 - \frac{1}{3^3}\right) \cdots \left(1 - \frac{1}{3^n}\right)\\[6pt] >{}& \left(1 - \frac13 - \frac{1}{3^2} - \frac{1}{3^3}\right)\left(1 - \frac{1}{3^4}\right) \cdots \left(1 - \frac{1}{3^n}\right)\\[6pt] >{}& \cdots\\ >{}& 1 - \frac13 - \frac{1}{3^2}- \frac{1}{3^3} - \cdots - \frac{1}{3^n}\\[6pt] ={}& \frac12 + \frac{1}{2\cdot 3^n}\\ >{}& \frac12. \end{align*}

Answer 2

The Euler function takes two forms: $$\phi(q):=\prod_{k=1}^\infty (1-q^k)=\sum_{k=-\infty}^\infty(-1)^kq^{(3k^2-k)/2},\quad\text{for }|q|<1.$$ The second one provides the desired lower bound: $$\begin{align}\prod_{k=1}^n\left(1-\frac1{3^k}\right)&>\phi\left(\frac13\right)\\&>\sum_{k=-1}^1\frac{(-1)^k}{3^{(3k^2-k)/2}}\\&=-\frac1{3^2}+1-\frac13\\&=\frac59\\&>\frac12. \end{align}$$

A more elementary method is to use a lower bound for $\ln:$ $$\forall x\in[0,1/2]\quad\ln(1-x)\ge-x-x^2$$ $$\begin{align}\ln\prod_{k=1}^\infty\left(1-\frac1{3^k}\right)&=\sum_{k=1}^\infty\ln\left(1-\frac1{3^k}\right)\\&\ge\sum_{k=1}^\infty\left(-\frac1{3^k}-\frac1{3^{2k}}\right)\\&=-\frac12-\frac18=-\frac58 \end{align}$$ and $e^{-5/8}>0.53.$