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Trying to solve this question, it appears to be important to know the value of an infinite product

$$\displaystyle{\prod_{k=2}^\infty \left(1-\dfrac{1}{4^k}\right)}$$ which terms looks a lot like a "geometric series".

My question is: Is it possible to calculate the value of the product above?

More generally, Is there a general method to solve this kinds of infinite products? Is there a theory of infinite products that is more or less similar to the theory of infinite sums (i.e., series)? Where can I find it?

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    $\begingroup$ What do you mean by general theory? Do you mean to establish convergence? There is not really one general theory for evaluating products (or sums either). What we have is a wide set of methods that one can try to apply like for example 1) taking $\log$'s to get a sum we can try to solve 2) look for patterns in partial products and prove via induction 3) take advantage of known products and try to manipulate a given product to that form (Weierstrass products for complex functions etc.) and many more. Take a look at some of the product questions on this site and you will find many examples. $\endgroup$
    – Winther
    Sep 16, 2016 at 12:21
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    $\begingroup$ Not in terms of simple elementary functions. You can write it in terms of the q-Pockhammer symbol as $\prod_{k=1}^\infty\left(1 - \frac{1}{4^k}\right) = (1/4)_{1/4}$. $\endgroup$
    – Winther
    Sep 16, 2016 at 12:23
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    $\begingroup$ en.wikipedia.org/wiki/Euler_function $\endgroup$
    – polfosol
    Sep 16, 2016 at 12:26
  • $\begingroup$ I don't know if the Chapter 7 of matem.unam.mx/ernesto/LIBROS/AC/… could be helpful. $\endgroup$ Sep 16, 2016 at 13:58
  • $\begingroup$ @Winther what I meant with a general theory was something like different criteria to study the convergence of an infinite products. $\endgroup$
    – Darío G
    Sep 16, 2016 at 15:02

1 Answer 1

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The convergence of an infinite product is surprisingly more easy than you think:

$$\prod_{k=1}^\infty\left(1-\frac1{4^k}\right)\text{ converges iff }\log\left[\prod_{k=1}^\infty\left(1-\frac1{4^k}\right)\right]\text{ converges}$$

$$\log\left[\prod_{k=1}^\infty\left(1-\frac1{4^k}\right)\right]=\sum_{k=1}^\infty\log\left(1-\frac1{4^k}\right)$$

And from there, it becomes an infinite sum/series problem. By the ratio test:

$$\lim_{n\to\infty}\left|\frac{\log\left(1-\frac1{4^{n+1}}\right)}{\log\left(1-\frac1{4^n}\right)}\right|=\lim_{n\to\infty}\left|\frac{4^n\left(1-\frac1{4^n}\right)}{4^{n+1}\left(1-\frac1{4^{n+1}}\right)}\right|=\frac14$$

Thus, it converges.

Nicely, according to WolframAlpha, we have

$$\sum_{k=1}^\infty\log\left(1-\frac1{4^{n+1}}\right)\approx-0.3731854421599476447\dots$$

And so

$$\prod_{k=1}^\infty\left(1-\frac1{4^k}\right)=e^{-0.3731854421599476447\dots}=0.68853753711614810750717\dots$$

Of course, these are approximations, not closed forms, but approximations satisfy the problem of "calculating" the product.

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  • $\begingroup$ you are right. I just thought there was an easy formula as in the case of the geometric series, but now I see that probably is not that easy. However, as you point out, determine the convergence is easier. $\endgroup$
    – Darío G
    Sep 17, 2016 at 11:17
  • $\begingroup$ How did he get rid of the $\log$ in the ratio test? I don't see it $\endgroup$
    – Averroes2
    Oct 1, 2020 at 14:22
  • $\begingroup$ @Averroes2 Honestly can't remember myself, but you may use $\ln(1+x)\sim x$ as $x\to0$ to get $(-1/4^{n+1})/(-1/4^n)=1/4$. $\endgroup$ Oct 1, 2020 at 21:19
  • $\begingroup$ I think its L'hopital $\endgroup$
    – Averroes2
    Oct 2, 2020 at 4:18

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