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? 
$$\prod_{i=1}^{\infty} s_i$$
 A: 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.
A: See: "Dedekind eta function".
A: To elaborate, and extend on GEdgar's answer: there is what is called the $q$-Pochhammer symbol
$$(a;q)_n=\prod_{k=0}^{n-1} (1-aq^k)$$
and $(a;q)_\infty$ is interpreted straightforwardly. The product you are interested in is equivalent to $\left(\frac1{10};\frac1{10}\right)_\infty\approx0.8900100999989990000001$.
One can also express the $q$-Pochhammer symbol $(q;q)_\infty$ in terms of the Dedekind $\eta$ function $\eta(\tau)$ or the Jacobi $\vartheta$ function $\vartheta_2(z,q)$; in particular we have
$$\left(\frac1{10};\frac1{10}\right)_\infty=\sqrt[24]{10}\eta\left(\frac{i\log\,10}{2\pi}\right)=\frac{\sqrt[24]{10}}{\sqrt 3}\vartheta_2\left(\frac{\pi}{6},\frac1{\sqrt[6]{10}}\right)$$

I might as well... there is the following identity, due to Euler (the pentagonal number theorem):
$$(q;q)_\infty=\prod_{j=1}^\infty(1-q^j)=\sum_{k=-\infty}^\infty (-1)^k q^\frac{k(3k-1)}{2}$$
which, among other things, gives you a series you can use for quickly estimating your fine product:
$$\left(\frac1{10};\frac1{10}\right)_\infty=1+\sum_{k=1}^\infty (-1)^k\left(10^{-\frac{k}{2}(3k+1)}+10^{-\frac{k}{2}(3k-1)}\right)$$
Three terms of this series gives an approximation good to twenty digits; five terms of this series yields a fifty-digit approximation.
