twoshots wrote:
lau wrote:
twoshots wrote:
I'm not really sure I see at all why we need to invoke anything about the Riemann rearrangement theorem in order to resolve this. Obviously, regrouping is based on a naive assumption that ignores what it means for a series to converge, so it seems completely unnecessary to say that I transition through anything, pardon me if I'm still failing to grasp this. If I take,
(1-1)+(1-1)+..
and rewrite this as
1-1+1-1+1...
by dropping the parenthesis, I have already clearly constructed a series with ambiguous convergence. I mean,
Σ((-1)^i)
doesn't converge at all, so of course it doesn't converge absolutely, why bother including anything further?
My emboldening, above, answers your question.
But I'm not rearranging that series at all (at least not in the sense of the theorem in question). If we dubbed a_i = (1-1) then our series is Σa_i. Since according to the theorem in question (which I have, I think correctly, assumed nudel was talking about, although I'm not sure if you are needing to refer to it) rearrangement corresponds to permuting {a_i}, it follows that every a'_i ∈ {a_i} if the rearranged series is Σa'_i. Since 1 is an element of the sequence summed in the second rewritten series but not in the sequence summed for the first, it follows that the sequence summed in the second is clearly not a permutation of the first and hence that this series is not the result of a "rearrangement" per se.
What connection is there between your latest discussion of permutations, as applied to an isolated one of the two series you gave, where the elements of the main series are grouped in pairs, and the main, central series of your original "proof", which is the one where the fallacy is introduced?
http://mathworld.wolfram.com/RiemannSeriesTheorem.htmlQuote:
By a suitable rearrangement of terms, a conditionally convergent series may be made to converge to any desired value, or to diverge.
twoshots wrote:
Alrighty, speaking of bogus ways to prove 1=0
(1-1)+(1-1)+(1-1)+(1-1)+...=0+0+0...=0 Right?
Now, this obviuhsleh is equal to
1+(-1+1)+(-1+1)+(-1+1)+...=1+0+0+0...=1
∴0=1
Again, my emboldening, in your original post.
I assume that the word "obviously" is meant. The only way the step you have glossed over, with that word, is "obvious", is if the series without parentheses is being appealed to.
If that is not what you intended, please explain (with proof) the "obvious" process that you feel can be employed to move from your initial parenthesised series, to your final one.
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"Striking up conversations with strangers is an autistic person's version of extreme sports."
Kamran Nazeer
17 Jan 2009, 10:24 am
