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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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Paul Manta
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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 06 '12 at 11:24
  • Your shorthand doesn't make sense. $0.9 \cdot 0.99 \cdot 0.999 \cdot \ldots$ makes sense as denoting an infinite product. $0.9 \cdot 0.99 \cdot 0.999 \cdot \ldots \cdot 0.9999999999999$ makes sense as denoting a finite product. The one you wrote doesn't make sense. –  May 06 '12 at 11:25
  • @KVRaman Thanks! :) – Paul Manta May 06 '12 at 11:25
  • @Hurkyl I fixed the title. – Paul Manta May 06 '12 at 11:26
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    $$s_i = 1-\frac1{10^i}$$ – Asaf Karagila May 06 '12 at 11:28
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    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 06 '12 at 11:28
  • @PaulManta I changed the \cdots to \times because it was confusing with the numbers themselves with decimal values. – Kirthi Raman May 06 '12 at 11:28
  • @PaulManta this converges to $0.89$, but you have to think how to show that. – Kirthi Raman May 06 '12 at 11:32
  • @KVRaman in the product, i starts from 1 and not 0 – zubinmehta May 06 '12 at 11:43
  • This link provides a very decent approximation for your product. – TenaliRaman May 06 '12 at 11:44
  • @KV Raman: Pari/GP gives for 200 factors 0.8900100999989990000001000099999999899999 000000000010000009999999999998999999900000 000000000100000000999999999999999989999999 990000000000000000001000000000099999999999 9999999998999999999990000000000000... and this does not improve if we increase the number of factors. – Gottfried Helms May 06 '12 at 11:44
  • Wolfram|Alpha gives the following result. Interestingly, its decimal expansion seems to consist solely of $0$'s, $1$'s, $8$'s and $9$'s. It does not look periodic, but otherwise seems to be quite regular, so it might be possible to prove nice properties about it. – Dejan Govc May 06 '12 at 11:48
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    @Tenali, I don't see the relation between the product in your link and the one in the current question. – Gerry Myerson May 06 '12 at 12:44
  • @GerryMyerson Approximate $1 - x$ with $e^{-x}$ just like Robert does in that link. The product becomes $e^{\sum_{i = 1}^{\infty} -10^{-i}} = e^{-\frac{1}{9}} \sim 0.89$. Of course this approximation is not very great or anything. – TenaliRaman May 06 '12 at 13:42
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    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 06 '12 at 13:49
  • You can prove it converges because it is decreasing and bounded below by 0. – Ross Millikan May 06 '12 at 15:33

3 Answers3

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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.

J. M. ain't a mathematician
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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.

J. M. ain't a mathematician
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jmad
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See: "Dedekind eta function".

J. M. ain't a mathematician
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GEdgar
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