# Result of the product $0.9 \times 0.99 \times 0.999 \times ...$

My question has two parts:

1. 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: $$(s_i)_{i\ge1} = 1 - 10^{-i}$$

2. 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? $$\prod_{i=1}^{\infty} s_i$$

• How about $(1-\frac{1}{10})(1-\frac{1}{100})...= \prod_{i=0}^{\infty}(1-\frac{1}{10^i})$ May 6, 2012 at 11:24
• $$s_i = 1-\frac1{10^i}$$ May 6, 2012 at 11:28
• I've seen the corresponding product with powers of 2 instead of powers of 10 discussed somewhere. My recollection is that no simple expression in terms of well-known constants is known, nor is one expected, but I can't cite any references at the moment. May 6, 2012 at 11:28
• @Tenali, I don't see the relation between the product in your link and the one in the current question. May 6, 2012 at 12:44
• Note that the fact that your factors converge to 1 is necessary for the product to converge, but it is not sufficient. There are products of sequences that converge to 1 that do not converge. May 6, 2012 at 13:49

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

$$\left(\frac1{10};\frac1{10}\right)_\infty=\sqrt{10}\eta\left(\frac{i\log\,10}{2\pi}\right)=\frac{\sqrt{10}}{\sqrt 3}\vartheta_2\left(\frac{\pi}{6},\frac1{\sqrt{10}}\right)$$

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.

• No special function is too special for J.M.! :) May 7, 2012 at 4:13

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.

• «The pattern is so regular that I'm confident» are great last words! :D May 6, 2012 at 18:47
• Your formula follows from the Euler identity in J. M.'s answer. In particular, every two terms in J. M.'s summation correspond to one term in your summation. May 6, 2012 at 19:46
• @WillOrrick: indeed. Now that J.M.'s answer contains this neat expression, my formula is an (increasing, though) abomination!