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The book Irresistible Integrals by George Boros and Victor Moll on page 204 has the following identity

$\displaystyle \frac{1}{1+x}=\prod_{k=1}^{\infty}\left(\frac{k+x+1}{k+x} \times \frac{k}{k+1}\right)$

How does one derive this?

Thanks.

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Multiply out

$$\prod_{k=1}^N\left(\frac{k+x+1}{k+x} \times \frac{k}{k+1}\right)$$

and cancel like terms. You will be left with

$$\frac{1}{1+x}\left(\frac{N+x+1}{N+1}\right) .$$

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  • $\begingroup$ Assuming $x$ is an integer? $\endgroup$ – Alex B. Dec 9 '10 at 12:58
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    $\begingroup$ @Alex: The terms cancel whether or not $x$ is integer. $\endgroup$ – Derek Jennings Dec 9 '10 at 13:08
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    $\begingroup$ Sorry, I was being dense. $\endgroup$ – Alex B. Dec 9 '10 at 13:13
  • $\begingroup$ @Alex, Are you from future? why everything is dense in future? $\endgroup$ – Arjang Dec 22 '10 at 22:18
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Hint $ $ It telescopes $\, \rm\displaystyle \prod_{k\:=\:a}^{b} \frac{f(k\!+\!1)}{f(k)}\, = \ \frac{\color{green}{\rlap{---}f(a\!+\!1)}}{\color{#C00}{f(a)}}\frac{\color{royalblue}{\rlap{---}f(a\!+\!2)}}{\color{green}{\rlap{---}f(a\!+\!1)}}\frac{\phantom{\rlap{--}f(3)}}{\color{royalblue}{\rlap{---}f(a\!+\!2)}}\, \cdots\, \frac{\color{brown}{\rlap{--}f(b)}}{\phantom{\rlap{--}f(b)}}\frac{f(b\!+\!1)}{\color{brown}{\rlap{--}f(b)}} =\, \frac{f(b\!+\!1)}{\color{#c00}{f(a)}} $

Apply that to both factors in the product then take the limit as $\rm\ b\to\infty\:.$

For some other examples of additive/multiplicative telescopy see here or here or here.

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