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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 $n\geq 1$ ?

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Sure.

$$(3+2n)x_n=2nx_{n-1}\implies x_n=\frac{2n}{3+2n}x_{n-1}$$

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.

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    $\begingroup$ ... and this can be written in "closed form" using the Gamma function. $\endgroup$ – Robert Israel Apr 16 at 17:41
  • $\begingroup$ Is there a way to give a simple expression of the product ? $\endgroup$ – furyo Apr 17 at 7:30
  • $\begingroup$ What is "simple" for you? It's already in closed form. You can find some exotic functions to represent it in another way but it's not really going to get "simpler" in a meaningful way. $\endgroup$ – Tony S.F. Apr 17 at 8:30

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