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?



2 Answers 2


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

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

  • $\begingroup$ Assuming $x$ is an integer? $\endgroup$
    – Alex B.
    Dec 9, 2010 at 12:58
  • 1
    $\begingroup$ @Alex: The terms cancel whether or not $x$ is integer. $\endgroup$ Dec 9, 2010 at 13:08
  • 1
    $\begingroup$ Sorry, I was being dense. $\endgroup$
    – Alex B.
    Dec 9, 2010 at 13:13
  • $\begingroup$ @Alex, Are you from future? why everything is dense in future? $\endgroup$
    – jimjim
    Dec 22, 2010 at 22:18

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.