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$$\ln2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - 2(\frac{1}{2} + \frac{1}{4} + \cdots)$$ $$= (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) - (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots) = 0$$

thanks.

SchrodingersCat
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Jichao
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  • @Jichao: Something is wrong. how does $\frac{1}{3} - \frac{2}{3} =\frac{1}{3}$, the third term appears to be incorrect! –  Sep 05 '10 at 15:58
  • @Chandru1:Corrected and thanks for your latex corrections too. – Jichao Sep 05 '10 at 16:03
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    You could rewrite this as $\log2=\log2=\infty-2\infty=\infty-\infty=0$. – Robin Chapman Sep 05 '10 at 16:21
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    @Chandru1: Take a look again at what Robin wrote. He's replaced the sums of the divergent series in the question with $\infty$ to highlight the error in the reasoning. Rearranging in this way, we could make the sum "equal" any number we like. – Derek Jennings Sep 05 '10 at 19:34
  • The "infinity minus infinity" explanation is suspect, because the equations $\infty - 2 \infty = \infty - \infty = 0$ are (in this case) correct. In the claimed chain of equations Log(2)=A=B=C=0, what is true is that Log(2)=A > B = C= 0, but the subtraction of infinities argument challenges the correct step B=C. – T.. Sep 07 '10 at 03:13
  • @Derek Jennings: As T.. notes below, this is not an issue of rearranging the series. (The third expression is not obtained from the second by applying a permutation of the terms.) It is rather a problem of carelessy introducing the quantity $\infty$ into a calculation, which is what Robin Chapman's comment is alluding too. As is shown in T..'s answer below, if one is careful about precise rates at which the divergent series involved grow, the paradox disappears. – Matt E Sep 07 '10 at 03:40
  • @Matt E: How you can claim that no rearrangement is involved? The reasoning in the original statement is fallacious and not all the steps are shown. In fact, they are deliberately omitted to create the “paradox”. To claim that the conclusion was arrived at by one method over another is to claim to understand the mind of the person who wrote the “proof”; a rearrangement could easily have formed part of that “proof”. – Derek Jennings Sep 07 '10 at 17:14
  • @Derek Jennings: My understanding is that "rearrangement" of a series involves choosing a permutation $\pi: \mathbb N \to \mathbb N,$ and replacing $\sum_{i = 1}^{\infty} a_i$ by $\sum_{i = 1}^{\infty} a_{\pi(i)}.$ This is not what seems to be happening in the above series of equations, which rather involves adding and subtracting various terms. Perhaps I have too literal an interpretation of the expression "rearrangement"? – Matt E Sep 07 '10 at 17:19
  • @Matt E: I think the key word in your reply is "seems". Anyway, I believe we both understand the subtleties of each other's argument :-) – Derek Jennings Sep 07 '10 at 17:29
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    @Derek Jennings:I think there is no rearrangement too because there is no permutation. – Jichao Sep 08 '10 at 01:28
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    @Derek Jennings: the claim that because "infinities" (or divergent sums, or conditionally convergent sums) are manipulated, any sum can be obtained, is incorrect. One can extend the definition of equality of convergent sums to equality of arbitrary sums in a logically consistent way (introducing no new relations between convergent sums), and most equations of "infinite" sums in this paradox are correct in that interpretation. The false step does not use rearrangement in the sense of Riemann's theorem on conditionally convergent sums being rearranged (permuted) so as to have any desired sum. – T.. Sep 08 '10 at 04:51

3 Answers3

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It's because the series for ln2 is conditionally convergent. (see also Riemann's rearrangement theorem)

Matt Calhoun
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    There is no rearrangement involved in the paradoxical calculation. – T.. Sep 07 '10 at 01:33
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    I have downvoted this answer because, as T.. has pointed out, it doesn't really address the paradox. The problem is not one of rearrangement, but rather of carelessly introducing divergent series into a calculation. If one is, instead, careful about how these divergent series are introduced, as is explained in T..'s answer, then the paradox disappears. Robin Chapman's comment gives another explanation. – Matt E Sep 07 '10 at 03:17
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    The series is still conditionally convergent, even if there is not rearrangement involved. – Asaf Karagila Sep 07 '10 at 15:30
  • @Asaf: although the calculations of this type are correct for absolutely convergent series, they are also correct for some conditionally convergent series. The (alternating) sum of 1/n is exactly at the boundary between the series for which it works and those for which it fails. This means that a correct explanation of the problem has to include concepts beyond conditional convergence. – T.. Sep 07 '10 at 16:29
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    @T..: I'm looking at the first series equity, and you take an alternating sum, and turn it into a "sum all positive terms" minus "sum of all negative terms". It sure looks like a rearrangement of a convergent series to me, mister. If you'd like it to be legit, then you should write it as $1+\frac{1}{2}-1+\frac{1/3}+\ldots$ and not as it is written in the question. – Asaf Karagila Sep 07 '10 at 16:39
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    @Asaf: Yes, the series is conditionally convergent, so that removing the signs gives a divergence. But the manipulations in the question don't involve a permutation of the indices (i.e. a rearrangement), so the allusion to the rearrangment theorem as an explanation doesn't address the question. – Matt E Sep 07 '10 at 16:43
  • @Asaf: I just saw your latest comment: how do you see a rearrangment? For example, the term $+1/2$ appears in the rewritten expression, but does not appear in the original sum (which has $-1/2$ instead)? What am I missing? – Matt E Sep 07 '10 at 16:44
  • @Matt E: Correct me if I'm wrong, the term $-\frac{1}{2}$ is the combination of $+\frac{1}{2}$ as well $-2\frac{1}{2}$ this would require an infinite number of commutations, which is undefined in this case. And the rearrangement? Well, first you replaced $-\frac{1}{2}$ by $\frac{1}{2}-1$ (and so on for the negative terms) and then you rearranged it to be written like that. – Asaf Karagila Sep 07 '10 at 17:02
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    @Asaf: Dear Asaf, I think I have a more literal (perhaps too literal?) interpretation of the term "rearrangement". Namely, I thought that a rearrangement of a series meant choosing a permutation $\pi:\mathbb N \to \mathbb N$, and then replacing $\sum_{i = 0}^{\infty}$ by $\sum_{i=0}^{\infty} a_{\pi(i)}.$ So I agree with your interpretation of the original question, but wouldn't call this a literal rearrangement. (Also, I think that Riemann's theorem refers to a literal rearrangement of terms as in my interpretation, rather than to the more elaborate process involved here. Is that right?) – Matt E Sep 07 '10 at 17:28
  • Well, if you look at it this way, first you've replaced all the negative terms by pairs of positive and negative terms, which was legal and all that. Then you applied the permutation, moving all the negative terms to one side and all the positive terms to the other, then you've taken out the $-2$ from the negative terms. But there was still a rearrangement, which was illegal. I think I'll turn this whole discussion of ours into an answer. – Asaf Karagila Sep 07 '10 at 19:49
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    @Asaf: the first equation is not a "rearrangement" of the series (sum the same terms in another order), but what is called a "splitting" of the series. As I indicated above, there are conditionally convergent alternating series a(1) - a(2) + a(3) - a(4) +... that can be split as in the first equation, and in fact the whole chain of equations would hold for such series. So this is NOT a problem that is fully addressed by discussing conditional convergence. The specific rate of divergence for the harmonic series Sum(1/n) is what makes the reasoning false, not divergence alone. – T.. Sep 08 '10 at 00:38
  • @T..: I see, I didn't know that. Well, I'd guess that the splitting would require slightly more than just being conditionally convergent and probably that each part of the split would converge? – Asaf Karagila Sep 08 '10 at 00:55
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    The calculations with finite sums tell you what is true, since everything is just a formalism for manipulating the finite sums. Here we calculated that the error in the first equation is a discrepancy between S(2n) and S(n) where S is the partial sums of the harmonic series. For the calculation to work we need S(2n)-S(n) to go to zero, which is true (e.g.) for any series dominated by the harmonic series: a(n)/n tending to zero. The logarithmic growth of Sum(1/n) in this example, is exactly the rate of divergence at which the "paradoxical" calculation becomes false. – T.. Sep 08 '10 at 01:13
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    (a typo: in the condition for being termwise dominated by the harmonic series it is a(n)/(1/n) = n*a(n) that must tend to zero, not a(n)/n.) – T.. Sep 08 '10 at 04:58
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In this chain of 4 equations, #1, #3 and #4 are correct, and no.2 is the mistake. The equations are assertions about (limits of) some finite sums. Let $H_n = \Sigma_{j=1}^n 1/j$ and $A_n = \Sigma_{i=1}^n (-1)^{i-1} 1/i$.

The correct formula is $A_n = H_n - H_{[n/2]}$ (which is approximately $\log(n) - \log(n/2) \sim \log(2) = A_{\infty}$) , but the second equation cut off at $n$ terms is claiming $A_n = H_n - H_n$ (which is zero).

T..
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  • Could you explain why A_n = H_n - H_n/2A_n = H_n - H_{n/2}? – Jichao Sep 07 '10 at 15:07
  • It is from repeating the original calculation for the finite sums. For even indices, A_2n = 1 - 1/2 + ... + 1/(2n-1) - 1/2n = H_2n - 2(1/2 + 1/4 + ... + 1/2n) = H_2n - H_n , and for odd indices, A_(2n+1) = H_(2n+1) - H_n. The notation [x/2] refers to the "greatest integer" or "floor" function, (the largest integer that is at most x) so that the formula for A_n handles both cases. – T.. Sep 07 '10 at 15:35
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In Matt's answer I began discussing with Matt E over several comments, which I think should be written out as an answer.

As Matt pointed out, this is a rearrangement of this conditionally convergent series which is why you have this sort of paradox.

However it was unclear about how this is exactly a rearrangement, as the equities seems perfectly legal - even for a conditionally convergent series.

  1. $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots = 1 + (\frac{1}{2} - 1) + \frac{1}{3} + (\frac{1}{4} - \frac{1}{2})$ is the first step, which is legal as you simply replace the negative terms by pairs of a positive and negative terms, but you don't change the order of summation from the original series which makes this exchange legit.
  2. $1 + (\frac{1}{2} - 1) + \frac{1}{3} + (\frac{1}{4} - \frac{1}{2}) = 1 + \frac{1}{2} + \frac{1}{3} + \ldots - 1 - \frac{1}{2} - \ldots$ this is where things break, you've taken a conditionally convergent series and changed the order, basically we've performed infinitely many commutations in order to rearrange the series into this order, and that is what breaks the summation.

The rearrangement wasn't very obvious, but it was hiding there with its big sharp pointy teeth... and when you stepped too close to its cave - it jumped out at you and bit your head off.

The series in the question is closely reminding me of the one my calculus teacher used when he first showed us what changing conditionally convergent series can do, although his was even less obvious.

Asaf Karagila
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  • The first equation of the paradox writes (1 - 1/2 + 1/3 etc) = P - Q, where P and Q are infinite series. This has a logically consistent interpretation as manipulation in the vector space of all infinite sums (convergent or not), with equality meaning the difference converges to zero. Writing equations between divergent quantities in this way, does not (and cannot) create new equations between convergent series, such as the paradox here with log(2)=0. Divergent infinite sums in a calculation do not make it wrong. Here the problem is from re-indexing via 2a(2n)=a(n), where a(n)=1/n. – T.. Sep 08 '10 at 00:54
  • Dear Asaf, I agree that step 2 is where the problem is. The point of my remark was merely that this particular manipulation is not a rearrangement of the series, in the literal sense of applying a permutation to the indices. It is a different kind of rewriting of the series, which causes problems for reasons that are addressed in T..'s answers and various comments. – Matt E Sep 08 '10 at 04:54
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    Asaf, splitting one series into two series can be seen as a bijection from $\omega$ to $\omega + \omega$, whereas a rearrangement is from $\omega$ to itself. Splitting is certainly a form of reorganization of the series, but is not rearrangement in the usual sense. – T.. Sep 08 '10 at 05:16
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    And they say you can't learn anything new at 4am! Thanks guys. – Asaf Karagila Sep 08 '10 at 07:31