<s>[INST] How can I prove that $\prod\limits_{n=1}^{\infty}\left(1-\frac{1}{x_{n}^{3/2}}\right)^{-1}=\frac{3}{2}$ if $x_{n}=x_{n-1}+\sqrt{x_{n-1}}+1, x_{0}=1$? If
$$x_{n}=x_{n-1}+\sqrt{x_{n-1}}+1, x_{0}=1$$
so we can say, that
$$\prod\limits_{n=1}^{\infty}\left(1-\frac{1}{x_{n}^{3/2}}\right)^{-1}=\left[\left(1-\frac{1}{3^{3/2}}\right)\left(1-\frac{1}{(4+\sqrt{3})^{3/2}}\right)\left(1-\frac{1}{(5+\sqrt{3}+\sqrt{4+\sqrt{3}})^{3/2}}\right)\cdots\right]^{-1}=\frac{3}{2}$$
All square roots are positive.
How can I prove it?
 [/INST] Let $u_n = \sqrt{x_n}$, we have
$$u_n^2 = u_{n-1}^2 + u_{n-1} + 1 = \frac{u_{n-1}^3 - 1}{u_{n-1} - 1}
\implies u_{n-1}^3-1 = u_n^2(u_{n-1}-1)
$$
This leads to
$$1 - \frac{1}{x_n^{3/2}} 
= \frac{u_n^3-1}{u_n^3} 
= \frac{u_{n+1}^2(u_n-1)}{u_n^3}
= \frac{u_{n+1}^2(u_n^2-1)}{u_n^3(u_n+1)}
= \frac{u_{n+1}^2(u_n^2-1)}{u_n^2(u_{n+1}^2-1)}\\
= \left.\frac{x_n - 1}{x_n}\right/\frac{x_{n+1}-1}{x_{n+1}}
$$
The product at hand is a telescoping one with partial products
$$\prod_{n=1}^p \left(1 - \frac{1}{x_n^{3/2}}\right)^{-1}
=\left.\frac{x_1}{x_1 - 1}\right/\frac{x_{p+1}}{x_{p+1}-1}$$
Since $x_n = x_{n-1}+\sqrt{x_{n-1}} + 1 \ge x_{n-1} + 1$ and $x_0 = 1$, we can show $x_n \ge n + 1$ by induction. This means $$\lim\limits_{p\to\infty} x_p = \infty \implies \lim\limits_{p\to\infty}\frac{x_{p+1}}{x_{p+1}-1} = 1$$
As a result,
$$\prod_{n=1}^\infty \left(1 - \frac{1}{x_n^{3/2}}\right)^{-1} = \frac{x_1}{x_1 - 1} = \frac{3}{3-1} = \frac32$$
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