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$$S = 1 - \frac12 + \frac13 - \frac14 + \frac15 - \frac16 + \frac17 - \frac18 + \frac19 - \frac1{10} + \frac1{11} - \frac1{12}\ldots\text{(to infinity)}$$

Rearranged, this series looks like: $$S = \left(1 - \frac12\right) - \left(\frac14\right) + \left(\frac13 - \frac16\right) - \left(\frac18\right) + \left(\frac15 - \frac1{10}\right) - \left(\frac1{12}\right) + \left(\frac17 - \frac1{14}\right) \ldots\text{(to infinity)}\\ S = \left(\frac12\right) - \left(\frac14\right) + \left(\frac16\right) - \left(\frac18\right) + \left(\frac1{10}\right) - \left(\frac1{12}\right) + \left(\frac1{14}\right) \ldots\text{(to infinity)}$$

This rearranged infinite series contains every number that the original infinite series had.

Further,

$$ 2S = 1 - \frac12 + \frac13 - \frac14 + \frac15 - \frac16 + \frac17 - \frac18 + \frac19 - \frac1{10} + \frac1{11} - \frac1{12} \ldots\text{(to infinity)}$$ Thus: $2S = S$
$2 = 1$
Mathematics disproven. Sorry.

Jokes aside, I know that infinite series can be calculated in different ways to get different results. My question is: Why? While it makes sense with Grandi's and similar series, it doesn't make sense to me for a series whose final term is $\frac1\infty = 0$.

MJD
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Niobius
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    If the series is absolutely convergent, you can't do that, every rearrangement leaves the sum unchanged. If the series is convergent but not absolutely convergent, the sum of the positive terms is $+\infty$, and the sum of the negative terms is $-\infty$. That means you can rearrange it to get any sum you please with the same terms in different order. Roughly, allowing arbitrary rearrangements, you get $\infty - \infty = \text{ undefined}$. – Daniel Fischer Jan 16 '14 at 14:12
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    You have to show $S\neq 0$ – Giulio Bresciani Jan 16 '14 at 14:17
  • @GiulioBresciani Why? The first series has sum $\ln(2)$. – apnorton Jan 16 '14 at 14:18
  • @anorton Yes, I know that $S=\ln(2)\neq 0$, but Niobus "proof" doesn't mention it, so it's wrong not only where he purposely made an error. I need a "sarcasm" sign – Giulio Bresciani Jan 16 '14 at 14:30

3 Answers3

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This is a good opportunity to point out something that many people don't get, or at least get accustomed to glossing over. When you write something like $S = 1 - 1/2 + 1/3 - 1/4 + 1/5$, you really are saying that $S$ is a sum; you can explicitly add the terms (and get $47/60$ in this specific case).

On the other hand when you write something like $S = 1 - 1/2 + 1/3 - 1/4 + 1/5 - \cdots$, this is a completely different animal. You can't explicitly add the terms because there are infinitely many of them. $S$ is not a sum, even though it is traditional to say so (I call it a sum as well). It may be spoken of as an "infinite sum", and it's convenient to regard it that way be cause in the right conditions it acts like a sum, but it is not--it is actually a limit.

For each $n$, the partial sum $S_n = 1 - 1/2 + \cdots + 1/n$ is indeed a sum which can be computed by explicitly adding the terms. The "infinite sum" $S$ is not a sum at all, but rather the limit $\lim\limits_{n\rightarrow\infty}S_n$, if it exists.

Often it is fine to forget about that and pretend $S$ is really a sum, but that can get you in trouble sometimes. So always keep in the back of your mind the notion that infinite sums are really limits of finite sums.

MPW
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    So what is the proof that $S \neq 2S$? Is it just because S is a limit and not a sum? Where exactly lies the "error"? – Jason Mar 12 '18 at 16:02
  • @Jason: First of all, writing S = anything with "..." in it is meaningless if the implied limit doesn't exist (and it doesn't in this case). By "implied limit" I mean the value of the expression with "..." in it. Since $S$ is not even a well-defined number, it certainly doesn't make sense to say $S=2S$ as neither side is a number. At the very best, the series might diverge to infinity, in which case you could reasonably state that "$\infty = 2\cdot\infty$", although one doesn't normally use arithmetic with infinity unless it is very carefully defined. – MPW Mar 12 '18 at 16:14
  • @Jason: (continued) Furthermore, even if it is true that $S=2S$, you can't cancel because this actually implies $S=0$ (just subtract $S$ from both sides to see that). After all, it is certainly true that $1\cdot 0 = 2\cdot 0$, but that doesn't mean you can cancel to conclude $1=2$. – MPW Mar 12 '18 at 16:16
  • Why are we saying that the implied limit doesn't exist? Isn't it ln(2)? So if we re-arrange the terms, we get 2ln(2). So how is this possible? – Sal.Cognato Mar 13 '18 at 19:43
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    @Sal.Cognato : Yes, you're right. I answered this 2 yrs ago and didn't reread carefully. In the end, the error is rearranging terms in a conditionally convergent series. The limit of the corresponding absolute series is what does not exist. – MPW Mar 13 '18 at 21:41
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If a series $$\sum_{i=0}^\infty a_i$$ is absolutely convergent, i.e. the sum $$\sum_{i=0}^\infty |a_i|$$ is convergent, then it can be shown that for any permutation (bijective function) of natural numbers $\pi$, the sum $$\sum_{i=0}^\infty a_{\pi(i)}$$ is the same.

For a series which does not absolutely converge, the same proof does not work. In fact, if the series is conditionaly convergent (it converges, but not absolutely), you can prove that you can choose ANY real number and can find a permutation $\pi$ such chat the sum will converge toward that number. The way I understand is that for non-absolutely-convergent series, the sum is "just barely well defined". It's only "truly" well defined for absolutely convergent series.

5xum
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    What you write in the second paragraph is not true without further assumptions. You need to at least assume that it is conditionally convergent. – Samuel Jan 16 '14 at 14:30
  • While that's a nice way of thinking, I think a better way of interpreting the failure of rearrangement in the non-absolutely convergent case is as a reminder that the order of the series is a necessary part of the data. After all, if you change the order, you get a different sequence of partial sums, so in some sense it shouldn't be surprising that it might have a different limit. – mdp Jan 16 '14 at 14:31
  • Samuel: Thank you, correction noted

    @MattPressland: A matter of interpretation, I guess. For me, the fact that the order is important is what I (internaly) see as "bad" definedness-The sum is only defined if you also give the order in which to sum it. An absolutely convergent series is "better defined", since it converges no matter how you sum it-you only need the elements which you want to add up and you can calculate the sum.

     – 5xum Jan 16 '14 at 14:32
  • @5xum Certainly - I think both interpretations say something interesting. – mdp Jan 16 '14 at 14:38
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    I find the following generalization immensely interesting. If a series $\sum a_n$ with $a_n\in\mathbb{C}$ converges (say to $S$), then the set of all sums obtainable by convergent rearrangements is either (1) just the single point $S$, in which case the series is absolutely convergent; (2) a line through $S$; or (3) the whole plane $\mathbb{C}$. In the latter two cases, the series is conditionally convergent. Note that in the case that all of the terms are real, the line in (2) is just the real line. – MPW Jan 16 '14 at 15:06
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To get a formal argument you can look at a proof of the Riemann rearrangement theorem - I'll try to give a more intuitive explanation (although the proof I know of the theorem is based on the same idea).

Essentially, this happens when both the positive and negative terms of the series can become arbitrarily small, but the positive terms still sum to $+\infty$ and the negative ones to $-\infty$. A standard example of this is the alternating harmonic series $\sum_{n=1}^\infty(-1)^n\frac{1}{n}$; taking either only the positive terms or the negative terms gives a sequence which is (almost) just a scalar multiple of the harmonic series, so diverges, but the individual terms may become arbitrarily small.

Now, lets say we want to use the same collection of terms to sum to a particular value $x$. Let's assume $x$ is positive, although it doesn't really matter. We can start adding up positive terms until we get to something larger than $x$ (the positive terms sum to $\infty$, so we won't run out before this happens). Now we can start adding the negative terms to move back towards $x$ again, until we overshoot slightly. Then add positive terms, etc. The conditions ensure (although it takes some care to see this formally) that

a) you never run out of terms of the right sign and get stuck on one side of $x$ or the other, and

b) that the terms you are using get smaller, so the error after each bit of adding becomes smaller in the long run.

mdp
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