How to calculate a bound for this product?

Question

Consider the following product:

$$ \prod_{i=1..n} {\left(1 - {1 \over 2^i}\right)} $$

A numeric calculation, up to $n=20$, gives $0.288788370496567$. But how can I calculate its limit when $n$ goes to infinity?

Alternatively, how can I prove that, for every $n$, the product is larger than $0.25$ (or some larger constant)?

@labbhattacharjee: I was trying that, but if you want the exact solution you don't want to approximate $\ln(1-x)$ with a Taylor series. — Daniel Robert-Nicoud, Jun 15 '13 at 19:15
Have a look here: http://math.stackexchange.com/questions/141705/result-of-the-product-0-9-times-0-99-times-0-999-times?rq=1 — MichalisN, Jun 15 '13 at 19:26
To add to Michalis fine link containing J.M.'s excellent answer, jmad's link to Euler function and GEdgar's to Pentagonal number theorem let's add this reference to Finch's book on constants ($Q$ in page 356-7) and the entry A015083 of OEIS. — Raymond Manzoni, Jun 15 '13 at 20:13
I now see that this question has already been answered here: http://math.stackexchange.com/questions/258067/how-can-one-compute-this-simple-infinite-product and here: http://math.stackexchange.com/questions/3776/limit-of-a-particular-variety-of-infinite-product-series — Erel Segal-Halevi, Jun 16 '13 at 07:05
To summarize: by the Pentagonal number theorem, $(1-x)(1-x^2)... = 1-x-x^2+x^5+x^7-...$, and for $x=1/2$, taking the first 4 elements, it is easy to see that the product is at least 0.28. — Erel Segal-Halevi, Jun 16 '13 at 07:06

score 0 · Answer 1 · answered Jun 15 '13 at 20:06

Hint:

$$\prod_{k=1}^n 1-{1\over 2^k} = {1\over2}\left(\prod_{k=1}^{n-1} 1-{1\over2} \left({1\over2}\right)^k\right) = {1\over2}\left ( \prod_{k=1}^{n-1}1-2^{-k-1} \right) \implies$$ $$\prod_{k=1}^\infty 1-{1\over 2^k} = {1\over2}\left (\prod_{k=1}^{\infty}1-2^{-k-1} \right)$$

How to calculate a bound for this product?

1 Answers1