10
$\begingroup$

I have a clueless friend who believes that

$$ \sum_{n=1}^\infty \frac{1}{2^n} $$

doesn't equal $1$ in the 'normal arithmetical sense'. He doesn't believe that this series flat out equals $$1$$ Is he correct?

  • 0
    Sorry, edit made.2012-05-24
  • 8
    What is the "normal arithmetical sense" in which we can interpret infinite series?2012-05-24
  • 0
    He says the fact that it converges only proves the existence of a least upper bound.2012-05-24
  • 7
    @Fried: $\sum \frac{1}{2^n}$ is a symbol. It _denotes that least upper bound._2012-05-24
  • 0
    Does you friend believe that irrational numbers exist? If he believes they exist, then he has no other option than to accept that $\sum_{n=1}^{\infty} \frac1{2^n} = 1$.2012-05-24
  • 4
    @Marvis: I don't understand your reasoning.2012-05-24
  • 0
    It also has every number $<1$ as an eventual lower bound. The only number $\le 1$ and $>1-\epsilon$ for every $\epsilon>0$ is $1$.2012-05-24
  • 3
    Something that should be pointed out is that the "sum" from 1 to infinity does not converge. It is indeed itself the limit of a sequence of partial sums, so instead of converging to something, it equals something.2012-05-24
  • 1
    @Marvis: Irrational numbers are not relevant here, since this particular sum turns out to be rational. If you want to stay in the field of rational numbers, you can still define infinite series in the usual way, it's just more complicated to decide whether such a series converges (to a rational number).2012-05-24
  • 0
    @QiaochuYuan I just wanted to point out that a finitist won't accept the existence of irrationals and in general the concept of limits etc. The comment I made was probably misleading.2012-05-24
  • 4
    The point is that since there is no way to sum an infinite collection of nonzero numbers one-by-one, the meaning we ascribe to $\sum_{n=1}^\infty a_n$ is the limit of the partial sums, if that limit exists. You might not call this "the normal arithmetical sense", but then it's up to you to say what (if anything) is "the normal arithmetical sense" for such a series.2012-05-24
  • 7
    possible duplicate of [Does .99999... = 1?](http://math.stackexchange.com/questions/11/does-99999-1)2012-05-24
  • 0
    @RobertIsrael I just wanted to point out that a finitist won't accept the existence of irrationals and in general the concept of limits etc. The comment I initially made was probably misleading.2012-05-24
  • 0
    This question would make more sense to me if you meant to write $\sum_{n=0}^{\infty} 2^n = -1$ instead of $\sum_{n=1}^{\infty} 1/2^n = 1$. In that case the series really doesn't "equal" $-1$ in the "normal arithmetical sense", but we can assign it that value for one reason or another. Is this what your friend was talking about?2012-05-25
  • 1
    I think Marvis's comment is psychologically apt. People who have trouble accepting $0.99999\ldots=1$, or $0.11111\ldots=1$ in base 2, are not being consistent unless they are similarly uncomfortable with $0.3333\ldots=1/3$ or (perhaps even more so) with $3.14159\ldots=\pi$. Maybe they are - I have not taken a poll of people who raise this issue - but if so, it is curious that the question always revolves around $0.9999\ldots$ and never around $0.3333\ldots$ or $3.14159\ldots$. The issue is, of course, that there is a lot more to the construction of the real number system than meets the eye.2012-05-25
  • 0
    It has an infinity sign in it. 'normal arithmetical sense' does not apply.2014-05-31

3 Answers 3