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How can we prove or disprove this:

$$\begin{align} \prod_{n=1}^{\infty}\frac {2n}{2n-1}=\frac {2\cdot 4\cdot 6\cdot 8\cdots}{1\cdot 3\cdot 5\cdot 7\cdots } &=\sqrt2\cdot \sqrt{\frac{2\cdot 4\cdot 4\cdot 6\cdot 6\cdot8\cdot8\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdot7\cdots}}\\ &=\sqrt{\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\cdot8\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdot7\cdots}}\\ &=\sqrt{\frac{\pi}2}&&\text{(using Wallis Product)} \end{align}$$

According to Wolframlpha, the product diverges, and the partial product is given by $$\prod_{n=1}^m\frac {2n}{2n-1}=\sqrt{\pi}\frac {\Gamma(m+1)}{\Gamma(m+\frac 12)}$$

The main objective of this question is ascertain why the Wallis product cannot be used in deriving the solution.

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Every terms count \begin{align*} \frac{2 \cdot 4}{1\cdot 3} &= \sqrt{5} \times \sqrt{\frac{2\cdot 2 \cdot 4 \cdot 4}{1\cdot 3 \cdot 3 \cdot 5}} \\ \frac{2 \cdot 4 \cdot 6}{1\cdot 3 \cdot 5} &= \sqrt{7} \times \sqrt{\frac{2\cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdot 6} {1\cdot 3 \cdot 3 \cdot 5 \cdot 5\cdot 7}} \\ \frac{(2n))!!}{(2n-1)!!} &= \sqrt{2n+1} \times \sqrt{\prod_{k=1}^{n} \frac{4k^2}{4k^2-1}} \\ & \sim \sqrt{n\pi} \end{align*}

which blows up!!

  • 0
    Please explain the reason that $\prod_{n=1}^{\infty} \frac{4n^{2}}{4n^{2} - 1}$ is asymptotic to $\frac{\pi}{2}$. – 2017-02-01
  • 1
    That's the finte Wallis product up to $2n$ factors. More precisely, $$\frac{\pi}{2} \sim \left( 1+\frac{1}{4n}-\ldots \right) \prod_{k=1}^{n} \frac{4k^2}{4k^2-1}$$ – 2017-02-01
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    Thanks - nice answer (+1). – 2017-02-02