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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.

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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.2010-09-08

3 Answers 3

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

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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.)2010-09-08
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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).

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    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.2010-09-07
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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.

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    And they say you can't learn anything new at 4am! Thanks guys.2010-09-08