Evaluating $\frac 9 {10}\cdot\frac {99} {100}\cdot\frac {999} {1000}\cdots$ 
$\displaystyle\frac 9 {10}\cdot\frac {99} {100}\cdot\frac {999} {1000}\cdots=?$

Usually, product of infinite many numbers which are less than 1, is 0. But How about this time?
Thank you.
 A: It's $\displaystyle \left(\frac{1}{10};\frac{1}{10}\right)_\infty=\sqrt[24]{10}\eta\left(-\frac{\log 10}{2\pi i}\right)\approx \href{http://www.wolframalpha.com/input/?i=%2810%5E1%2F24%29*DedekindEta%5B-+Log%5B10%5D+%2F+%282*Pi*I%29%5D}{0.890010099998999000000100009999999989999900}...$
See q-Pochhammer symbol and Dedekind eta function for details.
Proving that the product converges to a positive value can be proven elementarily. We have
$$\prod_{k=2}^n\left(1-\frac{1}{k^2}\right)=\left[\frac{\bf 2-1}{\color{Blue}{2}}\frac{\color{Blue}{3-1}}{\color{Red}{3}}\frac{\color{Red}{4-1}}{\color{DarkGreen}{4}}\cdots\frac{\color{DarkOrange}{n-1}}{\bf n}\right]\left[\frac{\color{Blue}{2+1}}{\bf 2}\frac{\color{Red}{3+1}}{\color{Blue}{3}}\frac{\color{DarkGreen}{4+1}}{\color{Red}{4}}\cdots\frac{\bf n+1}{\color{DarkOrange}{n}}\right]$$
$$=\frac{2-1}{n}\frac{n+1}{2}\longrightarrow \frac{1}{2}\quad\textrm{as}~n\to\infty.$$
By comparison, $k^2<10^{k-1}\implies 1-1/k^2<1-10^{-k}$, which shows our value converges to $>1/2$.
