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?

  • 1
    $\begingroup$ 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

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.


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.