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Does the sum of reciprocal of the numbers which digital expansion does not have an even amount of the digits $0-9$ converge? An example of a number that wouldn't be added is $\frac{1}{11}$ because $1$ shows up an even amount of times, same with $\frac{1}{1235412134}$ because $2,3,$ and $\space 4$ appear an even amount of times. The sum would be

$$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{12}+\frac{1}{13}+\dots+\frac{1}{1222}+\frac{1}{1234}+\dots$$

I think it would diverge because the elements are so close to the sum $\sum_{n=1}^{\infty} \frac{1}{n}$ which diverge, but I don't know how to prove it diverge, so it might converge.

  • We're not counting $0$ as even? – David C. Ullrich Jun 20 '21 at 21:14
  • we are not counting 0 –  Jun 20 '21 at 21:15
  • So having an even quantity of any digit means you do not include the reciprocal in the sum? So we would not include $\frac 1{11234}?$ – Ross Millikan Jun 20 '21 at 21:37
  • @RossMillikan yes we don't include $\frac{1}{11234}$ –  Jun 20 '21 at 21:50
  • It might be worth considering how many $n$ digit decimal integers do not have a digit appearing an even number of times. This could give you upper and lower bounds on the partial sums of the reciprocals up to $10^n$, and since the upper bound would be at worst $10$ times the lower bound, could tell you whether the original series converges or not – Henry Jun 20 '21 at 21:51
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    If for example there were at least $10^{2m-4}$ decimal integers between $10^{2m-2}$ and $10^{2m}-1$ not having a digit appearing an even number of times then, since their reciprocals would each be at least $\frac1{10^{2m}}$, your series would be at least as big as $\sum\limits_{m=1}^{\infty} \frac{10^{2m-4}}{10^{2m}} = \sum\limits_{m=1}^{\infty}10^{-4} = \infty$ – Henry Jun 20 '21 at 22:09
  • whose... (not suggesting an edit to avoid queue since it isn't substantial but still...) – Evariste Jun 20 '21 at 22:18
  • @Henry: the fraction is much smaller than that as there are $9$ different digits that each can appear an even number of times, but as long as you can bound the fraction we keep away from $0$ your argument goes through. I believe you can do that. – Ross Millikan Jun 20 '21 at 22:53
  • @RossMillikan My $10^{-4}$ term was me thinking that perhaps you had a probability of the order of $2^{-10}\approx 10^{-3}$ that all ten digits appeared an odd number of times, and a $0.81$ term to deal with the issue that the probability might be much lower for a number with an odd number of digits in total than one with an even number of digits plus the fact that you do not count much shorter numbers in this part of the calculation. I would be interested to any explicit example where the fraction was smaller. – Henry Jun 20 '21 at 23:08
  • @Henry: but if they are in the range of $10^{2m-2}$ you are only using a factor $10^{-2}$. I think it is similar to a typo. I think $10^{-4}$ is about right because of what you say. – Ross Millikan Jun 20 '21 at 23:11
  • @RossMillikan I am saying they are smaller than $10^{2m}$, but not smaller than $10^{2m-2}$ because I want to use those for a different part of the sum of reciprocals. So there are almost $10^{2m}$ of them to start with and my guess at least $10^{2m-4}$ which will satisfy the condition – Henry Jun 20 '21 at 23:24
  • @RossMillikan Further investigation suggests that $10^{-4}$ is pessimistic enough and that in fact, when $n$ is large and even, then $1.75781 \times 10^{n-3}$ is close to the number of $n$ digit integers with digits appearing an odd number of times (when $n$ is large and odd the number of satisfactory integers is substantially smaller, since ten odd numbers cannot add up to an odd number) – Henry Jun 21 '21 at 10:43
  • Very nice problem (+1) I have no idea for a rigorous proof yet, my guess would be "divergent" considering that the reciprocals of the primes of the form $an+b$ lead to a convergent series whenever $a$ and $b$ are coprime, but I admit this is only an intuition. – Peter Jun 22 '21 at 14:03
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    Why isn't $\frac 1{1230}$ included in the sum? It doesn't have an even number of each digit? Or by the first comment on this thread, do you propose to remove all numbers which contain $0$ as a digit? But then I see $\frac 1{10}$ : this might be a typo, but I'd like clarification anyway. Then again, I see the OP doesn't have an account. – Sarvesh Ravichandran Iyer Jun 22 '21 at 15:00

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