6
$\begingroup$

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

thanks.

  • 0
    @Jichao: Something is wrong. how does $\frac{1}{3} - \frac{2}{3} =\frac{1}{3}$, the third term appears to be incorrect!2010-09-05
  • 0
    @Chandru1:Corrected and thanks for your latex corrections too.2010-09-05
  • 5
    You could rewrite this as $\log2=\log2=\infty-2\infty=\infty-\infty=0$.2010-09-05
  • 2
    @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.2010-09-05
  • 0
    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.2010-09-07
  • 0
    @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.2010-09-07
  • 0
    @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”.2010-09-07
  • 0
    @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"?2010-09-07
  • 0
    @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 :-)2010-09-07
  • 1
    @Derek Jennings:I think there is no rearrangement too because there is no permutation.2010-09-08
  • 1
    @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