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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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2 Answers 2

up vote 9 down vote accepted

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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Assuming $x$ is an integer? –  Alex B. Dec 9 '10 at 12:58
1  
@Alex: The terms cancel whether or not $x$ is integer. –  Derek Jennings Dec 9 '10 at 13:08
1  
Sorry, I was being dense. –  Alex B. Dec 9 '10 at 13:13
    
@Alex, Are you from future? why everything is dense in future? –  Arjang Dec 22 '10 at 22:18

HINT $\ $ It telescopes: $\rm\displaystyle\ \ \prod_{k=a}^b\ \frac{f(k+1)}{f(k)}\ \ = \ \ \frac{\rlap{----}f(a+1)}{f(a)}\frac{\rlap{----}f(a+2)}{\rlap{----}f(a+1)}\ \ \cdots\ \ \frac{f(b+1)}{\rlap{--}f(b)}\ =\ \frac{f(b+1)}{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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