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# ValueofaVieta-likeproductsGeneralizedVieta-product

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It's known that
$$S_2={2\over\pi} = {\sqrt{2}\over 2}{\sqrt{2+\sqrt{2}}\over 2}{\sqrt{2+\sqrt{2+\sqrt{2}}}\over 2}\dots$$

The terms in the product approaches 1, the same holds for the following convergent series, with $\phi$ the golden ratio

$$S_1 = {\sqrt{1}\over\phi}{\sqrt{1+\sqrt{1}}\over\phi}{\sqrt{1+\sqrt{1+\sqrt{1}}}\over\phi}\dots$$

Is there a simpler formula for $S_n$?

Let $$S_n = {\sqrt{n}\over c_n}{\sqrt{n+\sqrt{n}}\over c_n}\dots$$

Where $c_n$ is the solution to the equation $x=\sqrt{n+x}$

Is it possible to extend there a simpler formula for $S_n$ to the positive reals? S_n$? What is the asymptotics asymptotic behavior (Big-O) of$S(n)$?S_n$ as $n->\infty$?

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It's known that
$$S_2={2\over\pi} = {\sqrt{2}\over 2}{\sqrt{2+\sqrt{2}}\over 2}{\sqrt{2+\sqrt{2+\sqrt{2}}}\over 2}\dots$$

The terms in the product approaches 1, the same holds for the following convergent series, with $\phi$ the golden ratio

$$S_1 = {\sqrt{1}\over\phi}{\sqrt{1+\sqrt{1}}\over\phi}{\sqrt{1+\sqrt{1+\sqrt{1}}}\over\phi}\dots$$

Is there a simpler formula for $S_n$?

$$S_n = {\sqrt{n}\over c_n}{\sqrt{n+\sqrt{n}}\over c_n}\dots$$

Where $c_n$ is the solution to the equation $x=\sqrt{n+x}$

Is it possible to extend $S_n$ to the positive reals? What is the asymptotics of $S(n)$ ?

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