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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

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

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 there a simpler formula for $S_n$?

What is the asymptotic behavior (Big-O) of $S_n$ as $n->\infty$?

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3  
the tag is wrong; please fix. this question was originally asked here: math.stackexchange.com/questions/6234/… –  Suvrit Apr 22 '11 at 14:00
    
Wait! What is $S_n$? –  Thierry Zell Apr 23 '11 at 15:02
    
@Suvrit: robotic's question is definitely related to mine, but different from it; I was curious about the behavior of $S_1$ specifically, whereas zie seems more curious about the asymptotics. –  Steven Stadnicki Apr 28 '11 at 21:50

1 Answer 1

up vote 6 down vote accepted

I doubt there exists a closed formula for $n\ne 2$. In the case $n=2$ such formula exists only thanks to the double-angle formula for cosine.

Let $n$ be fixed and $c=c_n$. Notice that $n=c^2-c$ and $c\to\infty$ as soon as $n\to\infty$.

Denote by $p_k$ the $k$-th multiplier in the product $S_n$. It can be easily seen that $$c\cdot (p_k^2-1) = p_{k-1} - 1$$

Consider the functional equation: $$c\cdot(f(x)^2-1)=f(2cx)-1$$ with $f(0)=1$ and $f'(0)=1$. Its solution can be expressed as a series: $$f(x) = 1 + x + \frac{x^2}{2(2c-1)} + \frac{x^3}{2(2c-1)^2(2c+1)} + \frac{x^4(2c+5)}{8(2c-1)^3(2c+1)(4c^2+2c+1)} + \dots.$$ Then $$p_k = f\left(\frac{x_0}{(2c)^k}\right)$$ where $x_0$ is a solution to $f(x_0)=0$.

Now $$S_n = \prod_{k=1}^{\infty} p_k = \exp \sum_{k=1}^{\infty} \ln\left(1 + \Theta\left(\frac{x_0}{(2c)^k}\right) \right) = \exp \sum_{k=1}^{\infty} \Theta\left(\frac{x_0}{(2c)^k}\right) = \exp \Theta\left(\frac{x_0}{2c-1}\right)$$ which tends to $1$ as $c\to\infty$.

Therefore, $S_n\to 1$ as $n\to\infty$.

Example. For $n=2$, the functional equation admits the analytic solution $f(x)=\cosh(\sqrt{2x})$ for which $x_0=\frac{-\pi^2}{8}$.

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This makes sense - so the behavior is $e^{Cn^{-1/2}}$ as $n\rightarrow\infty$? –  Steven Stadnicki Apr 28 '11 at 21:53
    
Yes. Moreover, $C=x_0/2$. –  Max Alekseyev Apr 29 '11 at 8:29

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