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To start out, I can demonstrate that any series of real numbers has a subseries that converges to an irrational.

Proof: Consider a sequence of positive real numbers, $a_n$, for whom $\sum a_n \in \mathbb R$. Because the sum of $a_n$ converges to a real then $a_n\to0$. Now we can take a subsequence of $a_n$ that converges as quickly as we like, so I'm going to find some subsequence of $a_n$, $b_n$, that has the property that $3b_{n+1} \le b_n$. Now every subsequence of $b_n$ has a 'unique' sum, in that no other subsequence of $b_n$ has the same sum as it. There are an uncountable number of subsequnces of $b_n$ and a countable number of rationals, thus some subsequence $b_n$, which in turn is a subsequence of $a_n$, must have a sum that converges to an irrational.

Now I've shown that every series of positive real numbers has a subseries that converges to an irrational number. My question is: "Is there an algorithm that would take in any series of real positive numbers that converges and would find a subseries that converges to an irrational number, and if there is such an algorithm what is it?"

Benji Altman
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    I'd take a look at Liouville's Approximation Theorem. If your chosen numbers decay rapidly enough then their sum converges to a transcendental number. – lulu Dec 12 '17 at 14:53
  • I think you can modify your method to extract a subseries that converges to any number (rational or not) that's less than the sum of the series. Not sure whether this would count as an algorithm. – Ethan Bolker Dec 12 '17 at 14:54
  • Say it converges to $a>0$ and pick $b>0,b<a.$ Your idea should make a subsequence converging (sum) to $b.$ – coffeemath Dec 12 '17 at 14:54
  • @EthanBolker I can give you the sequence $a_n = (10, .5, .25, .125, \ldots)$. It's not hard to show that $\sum a_n = 11$ however there are no subsequences converging to $8$. – Benji Altman Dec 12 '17 at 14:57
  • @EthanBolker this seems too strong. Take a decreasing sequence of positive numbers the sum of which converges to $.1$ and prepend $1$ in front. The new series converges, to $1.1$, but no subsequence converges to $.75$. – lulu Dec 12 '17 at 14:57
  • @lulu I think the Liouville's Approximation Theorem is a good way to go, thanks for that idea. – Benji Altman Dec 13 '17 at 23:35
  • @Ethan there is a general result that characterizes the possible sets of values you can get. For some series you indeed get an interval. For others you get a set homeomorphic to the Cantor set. The other possibility is a mixture of the two, a "cantorval". Z. Nitecki has a paper with details in the Monthly: – Andrés E. Caicedo Dec 14 '17 at 01:54
  • @Ethan MR3418208. Nitecki, Zbigniew. Cantorvals and subsum sets of null sequences. Amer. Math. Monthly 122 (2015), no. 9, 862–870. (Also available at the author's page, here.) – Andrés E. Caicedo Dec 14 '17 at 01:57

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