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I asked the wrong question here, my fault :(

How does one see, using $n! = \prod_{k=1}^\infty \left(\frac{k+1}{k}\right)^n \frac{k}{k+n}$, that

$$\left(\frac{1}{2}\right)! = \prod_{k=1}^\infty \left(\frac{k+1}{k}\right)^{\tfrac{1}{2}}\left( \frac{k}{k+\tfrac{1}{2}}\right) = \tfrac{1}{2}\sqrt{2 \cdot \left(\frac{2 \cdot 2}{1 \cdot 3}\frac{4 \cdot 4}{3 \cdot 5}\dotsm \right)}? $$

In other words, derive

$$ \prod_{k=1}^{\infty}\left(\frac{(2k)^2}{(2k+1)(2k-1)}\right)$$

while manipulating

$$\prod_{k=1}^\infty \left(\frac{k+1}{k}\right)^{\tfrac{1}{2}}\left( \frac{k}{k+\tfrac{1}{2}}\right)$$

Note this is a way to 'derive' the Wallis product

$$ \tfrac{1}{2}\sqrt{2 \cdot \left(\frac{2 \cdot 2}{1 \cdot 3}\frac{4 \cdot 4}{3 \cdot 5}\dotsm \right)} = \frac{\sqrt{2 \tfrac{\pi}{2}}}{2} = \frac{\sqrt{ \pi}}{2}$$

It should follow, "after a bit of manipulation", but it just doesn't! I must be missing something!

Edit: I now think it is not possible directly!

Community
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bobby
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1 Answers1

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Looks like you need to show that

$$\prod_{k=1}^{\infty}\left(\frac{(2k)^2}{(2k+1)(2k-1)}\right) = \frac{\pi}{2}.$$

This Wikipedia article uses Euler's infinite product for the sine function:

$$\frac{\sin x}{x} = \prod_{k=1}^{\infty}\left(1 - \frac{x^2}{k^2 \pi^2}\right).$$

The result follows quickly by applying $x = \pi/2.$

John
  • 26,319
  • No the point is to derive that infinite product from $\prod_{k=1}^\infty \left(\frac{k+1}{k}\right)^{\tfrac{1}{2}}\left( \frac{k}{k+\tfrac{1}{2}}\right)$ by a few manipulations! – bobby Dec 17 '14 at 23:21
  • I did the piece you needed at the end. (Well, Euler did. I just looked it up.) – John Dec 17 '14 at 23:25
  • I found an answer, a very cool one linked to above, in which we can use your step http://en.wikipedia.org/wiki/Wallis_product#Proof_using_Euler.27s_infinite_product_for_the_sine_function.5B1.5D to justify why the infinite product involves $\pi$! – bobby Dec 18 '14 at 00:22