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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

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

If instead of selecting terms you are willing to use two complete harmonic series at different rates, one positive and the other one negative, it is easy to get the logarithm of any positive rational.

$$ \log\left(\frac{p}{q}\right)=\sum_{i=0}^\infty \left(\sum_{j=pi+1}^{p(i+1)}\frac{1}{j}-\sum_{k=qi+1}^{q(i+1)}\frac{1}{k}\right) $$

For your irrational example $\pi$, something more elaborate is needed, but it is also possible. The idea is taking positive and negative terms in a ratio that tends to $e^\pi$. This may be achieved, for example, by taking advantage of this formula in setting non-constant length summations:

$$ e^\pi=e^{6asin\left(\frac{1}{2}\right)}=\sum_{k=0}^\infty\frac{3·2^{k-1}\left(e^{3\pi}-\left(-1\right)^ke^{-3\pi}\right)\Gamma\left(\frac{k}{2}+3i\right)\Gamma\left(\frac{k}{2}-3i\right)}{\pi2^kk!} $$

See also

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