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

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How about $(1-\frac{1}{10})(1-\frac{1}{100})...= \prod_{i=0}^{\infty}(1-\frac{1}{10^i})$ –  Kirthi Raman May 6 '12 at 11:24
$$s_i = 1-\frac1{10^i}$$ –  Asaf Karagila May 6 '12 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. –  Gerry Myerson May 6 '12 at 11:28
@Tenali, I don't see the relation between the product in your link and the one in the current question. –  Gerry Myerson May 6 '12 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. –  Phira May 6 '12 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

$$\left(\frac1{10};\frac1{10}\right)_\infty=\sqrt[24]{10}\eta\left(\frac{i\log\,10}{2\pi}\right)=\frac{\sqrt[24]{10}}{\sqrt 3}\vartheta_2\left(\frac{\pi}{6},\frac1{\sqrt[6]{10}}\right)$$

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.

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No special function is too special for J.M.! :) –  Bruno Joyal May 7 '12 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.

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«The pattern is so regular that I'm confident» are great last words! :D –  Mariano Suárez-Alvarez May 6 '12 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. –  Will Orrick May 6 '12 at 19:46
@WillOrrick: indeed. Now that J.M.'s answer contains this neat expression, my formula is an (increasing, though) abomination! –  jmad May 6 '12 at 20:26
Ramanujan style. –  dot dot Feb 26 '13 at 22:39
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Isn't the Euler function more relevant? –  jmad May 6 '12 at 12:51
And less well-known. You can't even call it Euler's function phi, since that is something else... –  GEdgar May 6 '12 at 13:26
For a series representation, see Euler's "pentagonal number theorem" ... en.wikipedia.org/wiki/Pentagonal_number_theorem –  GEdgar May 6 '12 at 13:33