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


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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! –  anonymous Sep 5 '10 at 15:58
You could rewrite this as $\log2=\log2=\infty-2\infty=\infty-\infty=0$. –  Robin Chapman Sep 5 '10 at 16:21
@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 5 '10 at 19:34
@Derek Jennings:I think there is no rearrangement too because there is no permutation. –  Jichao Sep 8 '10 at 1:28
@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 8 '10 at 4:51

3 Answers 3

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

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There is no rearrangement involved in the paradoxical calculation. –  T.. Sep 7 '10 at 1:33
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 7 '10 at 3:17
The series is still conditionally convergent, even if there is not rearrangement involved. –  Asaf Karagila Sep 7 '10 at 15:30
@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 7 '10 at 16:39
@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 7 '10 at 17:28

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).

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Could you explain why A_n = H_n - H_n/2A_n = H_n - H_{n/2}? –  Jichao Sep 7 '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 7 '10 at 15:35

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.

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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 8 '10 at 0:54
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 8 '10 at 4:54
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 8 '10 at 5:16
And they say you can't learn anything new at 4am! Thanks guys. –  Asaf Karagila Sep 8 '10 at 7:31

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