# $(3+2n)x_n = 2n x_{n-1}$

Is it possible to obtain a closed form expression of $$x_n$$ defined by $$x_0=2/3$$ and $$(3+2n)x_n = 2nx_{n-1}$$ for all $$n\geq 1$$ ?

$$(3+2n)x_n=2nx_{n-1}\implies x_n=\frac{2n}{3+2n}x_{n-1}$$
Since $$x_0=2/3$$, we can show by induction that $$x_n = \left(\frac{2}{3}\right)\prod\limits_{k=1}^n\frac{2k}{3+2k}$$
Recall that $$\prod\limits_{k=1}^n \frac{a_k}{b_k} = \frac{\prod\limits_{k=1}^n a_k}{\prod\limits_{k=1}^n b_k}$$. Using this we can also note that $$\prod\limits_{k=1}^n 2k = 2^n n! = (2n)!!$$ for the numerator and for the denominator we have $$(3 + 2)\cdot (3 + 4) \cdot (3 + 6) \cdot\ldots\cdot(3 + 2n)$$ which can be further reduced with the double factorial.