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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?

Later
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Friedmanite
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  • Sorry, edit made. – Friedmanite May 24 '12 at 20:40
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    What is the "normal arithmetical sense" in which we can interpret infinite series? – Chris Eagle May 24 '12 at 20:40
  • He says the fact that it converges only proves the existence of a least upper bound. – Friedmanite May 24 '12 at 20:42
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    @Fried: $\sum \frac{1}{2^n}$ is a symbol. It denotes that least upper bound. – Qiaochu Yuan May 24 '12 at 20:44
  • Does you friend believe that irrational numbers exist? If he believes they exist, then he has no other option than to accept that $\sum_{n=1}^{\infty} \frac1{2^n} = 1$. –  May 24 '12 at 20:44
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    @Marvis: I don't understand your reasoning. – Qiaochu Yuan May 24 '12 at 20:44
  • It also has every number $<1$ as an eventual lower bound. The only number $\le 1$ and $>1-\epsilon$ for every $\epsilon>0$ is $1$. – anon May 24 '12 at 20:44
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    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. – Tobias Kildetoft May 24 '12 at 20:47
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    @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). – Robert Israel May 24 '12 at 20:50
  • @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. –  May 24 '12 at 20:50
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    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. – Robert Israel May 24 '12 at 20:53
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    possible duplicate of Does .99999... = 1? – MJD May 24 '12 at 20:54
  • @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. –  May 24 '12 at 20:54
  • 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? – Antonio Vargas May 25 '12 at 14:41
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    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. – Will Orrick May 25 '12 at 15:00
  • It has an infinity sign in it. 'normal arithmetical sense' does not apply. – Christopher King May 31 '14 at 02:10

3 Answers3

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As far as I can tell, your friend is not distinguishing appropriately between a series and its sum. A series is a sequence of numbers $s_1, s_2, s_3, ...$ which one specifies by specifying $a_1 = s_1, a_n = s_n - s_{n-1}$. The notation $$\sum_{n=1}^{\infty} a_n$$

denotes the limit of the sequence $s_i$, if it exists, and is called the sum of the series. It is a number which is unique if it exists and should not be identified with the series itself.

Qiaochu Yuan
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If you want to give your friend a visual approach, try this. Draw a square. Bisect it vertically and fill in the left side (that corresponds to $1/2$). Then bisect the right rectangle horizontally, and fill in the bottom ($1/2+1/4$). Bisect the unfilled square vertically, and fill in the left ($1/2+1/4+1/8$). Continue on like this to give your friend the general idea.

The reason that it is equal to $1$ (i.e.: to the whole square) is that for every point inside the square, we can iterate this procedure far enough so that that point gets shaded over. In other words, this procedure fills in the square, taken to the limit, so the corresponding area (number) is at least $1$. But at every stage, we are only filling in sections inside the square, so it is at most $1$, too, and thus, equal to $1$.

Cameron Buie
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I had a question closed a while back about something very similar - write the sum in base 2. It comes out as 0.1111111 ... Is this the same as 1?

Well, yes, because that's how limits are defined. But my daughter was struggling towards some language about open sets and limit points (therefore closed sets) and wanted 0.999999 (use base 10) to be different from 1 to indicate (effectively) that the set of partial sums did not include the limit point.

I reckon that a 13-year-old who can even think of conceptualising that kind of thing (no suggestion from me) is doing pretty well. Especially as this is one of the harder conceptual leaps between what most students get at High School and what they have to deal with at university.

The question is mathematically resolved, but the resolution is much more subtle than the indoctrinated elite (like me) sometimes imagine.

Mark Bennet
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