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Let $a_n$ be a sequence of positive rational numbers who's sum is rational. I would like to construct a subsequence of $a_n$ who's sum is irrational. This is a special case of this question.

I know that such a number exists (see the generalized version for existance proof), and there was a suggestion to use Louiville's theorem.

Now I have an idea that might work but I'm not sure how to prove it or if it's correct. Let us start to construct our subsequence, we will start by including $a_1$. We will write $a_1$ in decimal and at some point the representation for $a_1$ will start repeating. For example it might be $.1483\ldots43\overline{748}$ or really anything else, but after some number of digits it will start repeating because it's rational. Lets say it starts repeating on the $k^{\text{th}}$ digit after the decimal, then we will pick the second element of our subsequence to be such that it is less then $10^{-k}$, and thus will 'mess' with the repeating part of the number. Now the sum of these first two things will have a repeating part later down in the number. We keep repeating this process and I think we end up with an irrational once we do this infinity.

Does this work as a construction? If so can you show why, and if not is there some other construction that will work?

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    $\begingroup$ You need to assume infinitely many $a_n$ are nonzero. $\endgroup$ – Robert Israel Dec 14 '17 at 1:17
  • $\begingroup$ @RobertIsrael You're right, I've edited to correct to what I meant to ask. $\endgroup$ – Benji Altman Dec 14 '17 at 2:15
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It is easy to create what seem like irrationals from subseries of rationals that converge to a rational. However you then you have the problem of proving the numbers you get this way are actually irrational. For example you could use the double series I created that always converges to a rational

$$\frac{z}{z-1}=\sum_{m=0}^\infty \sum_{k=1}^\infty \frac{1}{(k+z^m)\prod_{n=0}^{k-2}(n+z^m)}$$

See my question Rationals Approximated by a Fast Converging Series of Rationals

If keep $k$ constant you can create candidate irrationals, but how do you prove they are irrational? in the case $k=1$ and $z=2$ we have according to Mathematica

$$\sum_{m=0}^\infty \sum_{k=1}^1 \frac{1}{(k+2^m)\prod_{n=0}^{k-2}(n+2^m)}=\frac{-\log (2)+\psi _{\frac{1}{2}}^{(0)}\left(-\frac{i \pi }{\log (2)}\right)}{\log (2)}$$

In the general case for $k=1$ and $z$ we have according to Mathematica

$$\sum_{m=0}^\infty \sum_{k=1}^1 \frac{1}{(k+z^m)\prod_{n=0}^{k-2}(n+z^m)}=\frac{2 \psi _z^{(0)}\left(-\frac{i \pi }{\log (z)}\right)+2 \log (z-1)+\log (z)+2 i \pi }{2 \log (z)}$$

This is conjectural, but hopefully gives you some ideas. I don't know enough about the q-diagamma function have a good handle on how to prove these numbers are irrational.

There are wealth of other possibilities as well. This is the analogous double series for the zeta function

$$\zeta(s)=\sum_{m=0}^\infty \sum_{k=1}^\infty \frac{1}{(k+m^s)\prod_{n=0}^{k-2}(n+m^s)}$$

For $s=2$ and $k=1$ we have

$$\sum_{m=0}^\infty \sum_{k=1}^1 \frac{1}{(k+m^2)\prod_{n=0}^{k-2}(n+m^2)}=\frac{1}{2} (\pi \coth (\pi )-1)$$

I was trying to see if there is a difference between the two types of double series, one that converges to a rational and one that converges to an irrational, but I haven't made much headway on this as I have had other things to do.

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