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Given $\sum_{n=1}^{\infty} \frac{1}{2^n}$, what real numbers in $\left[ \frac{1}{2},1 \right]$ can I generate with subseries of this series?

Obviously we have every power of $\frac{1}{2^n}$ (by taking single terms), as well as 1 itself, which is the value of the original series. But can I get to any real number I want with the appropriate terms? Informally I would say yes, since we can approach with arbitrary precision by choosing the terms that are as small as I need.

Are all the reals in the interval $\left[ \frac{1}{2},1 \right]$ reachable? If not, is there a way to characterize "how many" are reachable?

Greg L
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    Hint: Base $2$. – Lord Soth Jul 15 '13 at 16:31
  • Every real number in [0.5,1] is of the form 0.XXX... where X is 0 or 1, and each digit corresponds to a term in this series? – Greg L Jul 15 '13 at 16:35
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    Let me mention a nice generalization: In The subset sum of a null sequence, Zbigniew Nitecki associates to each sequence $\vec x$ converging to $0$ the set of numbers that are the sum of a subsequence of $\vec x$, the subsequence may be infinite, finite, or even empty. He characterizes the possible shapes of this set. The example in this question is a bounded closed interval including $0$. – Andrés E. Caicedo Jul 15 '13 at 17:07
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    The possibilities are: An unbounded closed interval which includes zero, a finite union of (nontrivial) compact intervals, a Cantor set, or what he calls a “symmetric Cantorval”, that is, a nonempty compact subset $S$ of the real line such that (1) $S$ is the closure of its interior, and (2) Both endpoints of any nontrivial connected component of $S$ are accumulation points of trivial (i.e., one-point) connected components of $S$. – Andrés E. Caicedo Jul 15 '13 at 17:10
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    @AndresCaicedo, thanks for the reference but your link is broken, is http://www.tufts.edu/~znitecki/Subsum%20Sets%20(v9).pdf the correct paper? – Greg L Jul 15 '13 at 17:46
  • Ah, how annoying! Yes, that's the one. See also here. – Andrés E. Caicedo Jul 15 '13 at 17:54
  • Some of the results from the paper linked by Andres Caicedo were mention in this question. – Martin Sleziak Jul 15 '13 at 18:30
  • @MartinSleziak Thanks for the pointer. There are some nice references there. – Andrés E. Caicedo Jul 15 '13 at 21:07

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Just like any real number has a decimal expansion you can write any real number in base $2$. If you don't allow positive exponents or a minus sign you get all numbers between $0$ and $1$ (both included).

MichalisN
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  • $1$ is included. The series is a "subseries" of itself, and the sum of the series is $1$. – dfeuer Jul 15 '13 at 16:39
  • @dfeuer: Thanks, corrected – MichalisN Jul 15 '13 at 16:40
  • Thanks! I take it that this approach can also show that if we change 2 to any other natural number, we cannot find subseries that converge to arbitrary reals in the appropriate interval, because the numerator remains equal to 1 and so not every decimal expansion in the corresponding base can be recreated? – Greg L Jul 15 '13 at 16:44
  • Oof, of course the interval is [0,1], I feel stupid :) – Greg L Jul 15 '13 at 16:46
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    Yes, exactly. It only works with $2$ because the coefficients of the binary expansion are either $0$ (=omit term) or $1$ (=include term). – MichalisN Jul 15 '13 at 16:46
  • It does, however, work just fine for real numbers in $(1,2]$. Michalis's error was most likely a result of the fact that it's very common to consider base-$k$ representations that don't end in an infinite string of $0$s or ones that don't end in an infinite string of $k-1$s. That lets you get a unique representation of real numbers. – dfeuer Jul 15 '13 at 18:15