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What is the value of the infinite product $$ \frac{1}{2} \frac{3}{4} \frac{5}{6}\cdots = \prod_{n=1}^{\infty} \frac{2n-1}{2n} = \prod_{n=1}^{\infty} \left(1-\frac{1}{2n}\right) = \lim_{n\to\infty}\frac{(2n-1)!!}{(2 n)!!}$$

What is the asymptotic of the product as a function of $n$ for large $n$?

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  • $\begingroup$ @Ross Millikan You are right, corrected the formula $\endgroup$ Dec 20, 2017 at 5:25

2 Answers 2

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Without Stirling, but only comparison theorems for series with non-negative terms:

Letting $a_n \stackrel{\rm def}{=} 1-\frac{1}{2n}$, $$\begin{align*} \ln \prod_{n=1}^N a_n &= \sum_{n=1}^N \ln a_n = \sum_{n=1}^N \ln \left(1-\frac{1}{2n}\right) \operatorname*{=}_{N\to\infty} - \sum_{n=1}^N \frac{1}{2n}- \sum_{n=1}^N \underbrace{\left( \frac{1}{4n^2}+o\left(\frac{1}{n^2}\right)\right)}_{\stackrel{\rm def}{=} b_n} \\&= -\frac{1}{2}H_N - \sum_{n=1}^N b_n \tag{1} \end{align*}$$ using the fact that $\ln (1+x) \operatorname*{=}_{x\to0} x - \frac{x^2}{2}+o(x^2)$. Now, recall that $H_N \operatorname*{=}_{N\to\infty} \ln N + \gamma + o(1)$ to get $$\begin{align*} \ln \prod_{n=1}^N a_n &= -\frac{1}{2}\ln N - \frac{1}{2}\gamma +o(1) - \sum_{n=1}^N b_n\tag{2} \end{align*}$$ and, since $\sum_{n=1}^N b_n$ converges (by comparison with $\sum_n \frac{1}{n^2}$) to some real value (call it $c$), $$\begin{align*} \ln \prod_{n=1}^N a_n &= -\frac{1}{2}\ln N - \frac{1}{2}\gamma - c + o(1)\,.\tag{3} \end{align*}$$ By continuity of exponential, this gives" $$\begin{align*} \prod_{n=1}^N a_n &= \frac{1}{\sqrt{N}}\cdot e^{-\frac{1}{2}\gamma - c}\cdot e^{o(1)}\,.\tag{4} \end{align*}$$ and, since $\lim_{N\to\infty}e^o(1) = 1$, we get $$\begin{align*} \boxed{ \prod_{n=1}^N a_n \operatorname*{\sim}_{N\to\infty} \frac{C}{\sqrt{N}} } \end{align*}$$ for some positive constant $C\stackrel{\rm def}{=} e^{-\frac{1}{2}\gamma - c}$.

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    $\begingroup$ Note: that constant $C$ is actually (as found using other methods) equal to $\frac{1}{\sqrt{\pi}}$. $\endgroup$
    – Clement C.
    Dec 20, 2017 at 5:48
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Apply Stirling's Formula to $$\frac{(2n)!}{(2^nn!)^2}$$

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  • $\begingroup$ Very good answer, sterling formula $\lim_{x\rightarrow \infty} \log(N!)=N\log{N}-N$, $\endgroup$
    – Sonal_sqrt
    Dec 20, 2017 at 4:03
  • $\begingroup$ @PiyushDivyanakar: there is an additional $\frac 12 \log (2 \pi N)$. It doesn't look like it matters here. $\endgroup$ Dec 20, 2017 at 5:36

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