Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\pi$.

share|improve this question
add comment

2 Answers

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.

share|improve this answer
    
Aargh! The terms have to go to zero. I'll edit. –  Gerry Myerson Jul 15 '13 at 9:28
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.