Result of the product $0.9 \times 0.99 \times 0.999 \times ...$ My question has two parts:


*

*How can I nicely define the infinite sequence $0.9,\ 0.99,\ 0.999,\ \dots$? One option would be the recursive definition below; is there a nicer way to do this? Maybe put it in a form that makes the second question easier to answer. 
$$s_{i+1} = s_i + 9\cdot10^{-i-2},\ s_0 = 0.9$$ 
Edit: Suggested by Kirthi Raman:
$$(s_i)_{i\ge1} = 1 - 10^{-i}$$

*Once I have the sequence, what would be the limit of the infinite product below? I find the question interesting since $0.999... = 1$, so the product should converge (I think), but to what? What is the "last number" before $1$ (I know there is no such thing) that would contribute to the product? 
$$\prod_{i=1}^{\infty} s_i$$

By looking at the decimal representation, it appears that:
$$
\prod_{i=1}^\infty\left(1-\frac1{10^i}\right)=
\sum_{i=1}^\infty
  \frac{8 +
    \frac{10^{2^i-1}-1}{10^{2i-1}} +
    \frac1{10^{6i-2}} +
    \frac{10^{4i}-1}{10^{12i-2}}
  }{10^{(2i-1)(3i-2)}}
$$
I don't have a proof, but the pattern is so regular that I'm confident.
