Lower bound of product. We have the product :
$$\prod{(1-\frac{1}{2^i})}$$ I know that product has upper bound and possibly has lower bound.
My question is : how can I get the lower bound of this product?
My idea was to estimate this product with other product , but I didn't get the product with non-zero limit.
 A: Observe for $n\geq 4$ we have
\begin{align}
2^n \geq n^2
\end{align}
then it follows
\begin{align}
\prod^\infty_{n=4}\left(1-\frac{1}{n^2}\right) \leq \prod^\infty_{n=4}\left( 1- \frac{1}{2^n}\right)
\end{align}
which means
\begin{align}
\prod^\infty_{n=2}\left(1-\frac{1}{n^2}\right) \leq 
\frac{\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)}{\left(1-\frac{1}{2} \right)\left(1-\frac{1}{2^2} \right)\left(1-\frac{1}{2^3} \right)}
\prod^\infty_{n=1}\left( 1- \frac{1}{2^n}\right). 
\end{align}
Thus, we have
\begin{align}
\frac{63}{256}= \frac{63}{128}\prod^\infty_{n=2}\left( 1-\frac{1}{n^2} \right)\leq \prod^\infty_{n=1}\left( 1- \frac{1}{2^n}\right).
\end{align}
A: (very partial answer) 
Sequence $a_n:=\prod_{i=1}^n{(1-\frac{1}{2^i})}$ is positive (evident) and decreasing because $a_1=\frac{1}{2}$ and $a_{n}=(1-\frac{1}{2^n})a_{n-1}$. As it is bounded by below by $O$, it converges to its lower bound, by a classical result in analysis. 
In particular, it is the same thing to speak about "the" lower bound of sequence ($a_n$) and the "value" of the infinite product.
I don't know of a simple expression of this limit. Nevertheless, have a look to the recent following question which generalizes yours:(Power series for an infinite product.) (take $x=1/2$ and $z=1$) ; it shows how to convert this infinite product into a series.
