Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can all algebraic numbers (i.e. quantities such as $3/5$, $\sqrt{2}$, $\sqrt{3}$, etc.) be expressed as an infinite sum whose summands never permanently vanish?

A well known example is $$\sum_{n=1}^\infty\frac{1}{2^n}=1.$$

I realise we could always multiple this infinite summation by any number we like, but what if this were not permitted?

share|cite|improve this question
Then, exactly what numbers are supposed to be used in the summation? – Berci Nov 7 '12 at 10:46
What exactly are the rules? If $x \in \mathbb{R}$, then $x = -1 + x + \sum_{n=1}^{\infty} \frac{1}{2^n}$. – littleO Nov 7 '12 at 10:47
As @Berci says. You need to add more serious restrictions, or else the answer is trivially yes: Every real number is the sum of a series with rational summands, for example. – Harald Hanche-Olsen Nov 7 '12 at 10:48
Every real number, whether algebraic or otherwise can be represented thus: just use the decimal expansion, and chose the one which ends with all 9's from some point onwards if the decimal you chose happens to be a terminating one. – Old John Nov 7 '12 at 10:50

Your Answer


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

Browse other questions tagged or ask your own question.