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

  • 10
    $\begingroup$ How about $(1-\frac{1}{10})(1-\frac{1}{100})...= \prod_{i=0}^{\infty}(1-\frac{1}{10^i}) $ $\endgroup$ – Kirthi Raman May 6 '12 at 11:24
  • 3
    $\begingroup$ $$s_i = 1-\frac1{10^i}$$ $\endgroup$ – Asaf Karagila May 6 '12 at 11:28
  • 3
    $\begingroup$ 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. $\endgroup$ – Gerry Myerson May 6 '12 at 11:28
  • 1
    $\begingroup$ @Tenali, I don't see the relation between the product in your link and the one in the current question. $\endgroup$ – Gerry Myerson May 6 '12 at 12:44
  • 3
    $\begingroup$ 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. $\endgroup$ – 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.

  • 14
    $\begingroup$ No special function is too special for J.M.! :) $\endgroup$ – 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.

  • 14
    $\begingroup$ «The pattern is so regular that I'm confident» are great last words! :D $\endgroup$ – Mariano Suárez-Álvarez May 6 '12 at 18:47
  • 3
    $\begingroup$ 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. $\endgroup$ – Will Orrick May 6 '12 at 19:46
  • $\begingroup$ @WillOrrick: indeed. Now that J.M.'s answer contains this neat expression, my formula is an (increasing, though) abomination! $\endgroup$ – jmad May 6 '12 at 20:26
  • $\begingroup$ Ramanujan style. $\endgroup$ – dot dot Feb 26 '13 at 22:39

See: "Dedekind eta function".

  • $\begingroup$ Isn't the Euler function more relevant? $\endgroup$ – jmad May 6 '12 at 12:51
  • $\begingroup$ And less well-known. You can't even call it Euler's function phi, since that is something else... $\endgroup$ – GEdgar May 6 '12 at 13:26
  • $\begingroup$ For a series representation, see Euler's "pentagonal number theorem" ... en.wikipedia.org/wiki/Pentagonal_number_theorem $\endgroup$ – GEdgar May 6 '12 at 13:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.