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Suppose you take the set $X=\{\sum_{k \in A} \frac{1}{k}: A \in \mathcal{P}(\mathbb{N} \setminus \{1\})\}$. Suppose that we agree to introduce the symbol $\infty$ to encompass the cases where the series $\sum_{k \in A} \frac{1}{k}$ diverges (so $\infty \in X$). My question is if any irrational number (say, $\pi$) is in $X$.

Surely this could only possibly happen for an infinite set $A$ (any finite sum would have to be a rational number). Considering the fact that you can get converging series by deleting some of the terms of the harmonic series, it might happen that you could somehow obtain a series that converges to $\pi$.

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up vote 4 down vote accepted

You can get every positive number, rational or otherwise, as a sum of infinitely many reciprocals of distinct positive integers. That's true for any divergent series of positive terms [EDIT: as long as the terms go to zero]. I could tell you exactly how to do this, but you might get more out of working it out on your own.

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Aargh! The terms have to go to zero. I'll edit. –  Gerry Myerson Jul 15 '13 at 9:28
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It's well known that $\sum_{i=1}^{\infty}\frac{1}{i^2}=\frac{\pi^2}{6}$ by Euler, see Basel problem. And $\pi^2$ is irrational, see e.g. planetmath.

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Hmm true, we could take the terms that are in a $p$-series for any given natural $p$ greater than 1 and those terms would also be in the harmonic series, thanks. –  José Siqueira Jul 15 '13 at 10:35
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