# How to calculate a bound for this product?

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)?

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How about applying logarithm & then expand using $\ln(1-x)$ –  lab bhattacharjee Jun 15 '13 at 19:12
@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: math.stackexchange.com/questions/141705/… –  Michalis 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: math.stackexchange.com/questions/258067/… and here: math.stackexchange.com/questions/3776/… –  Erel Segal Halevi Jun 16 '13 at 7:05
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$$\prod_{k=1}^n 1-{1\over 2^k} = {1\over2}\left(\prod_{k=1}^{n-1} 1-{1\over2} \left({1\over2}\right)^k\right) = {1\over2}\left ( \prod_{k=1}^{n-1}1-2^{-k-1} \right) \implies$$ $$\prod_{k=1}^\infty 1-{1\over 2^k} = {1\over2}\left (\prod_{k=1}^{\infty}1-2^{-k-1} \right)$$