Why is it this a problem? My prof always shrug this off as something very delicate about the notion of infinity so unless we are careful we will be wrong. But this seems to be a philosophical answer more than a mathematical one.
Surely there's a well thought out reasoning using proper mathematics that explain what kind of rearrangement is acceptable.
Riemann only ever said you can add up converging alternating series to any value you want by rearranging right?
I am guessing this have to do with the fact that we have infinite amount of positive and negative term, so it is possible to do arrangement such that we can use it to approximate the summation to any value without "exhaustion" in a sense. Since our series is an convergent alternating series, our approximation will get better.
ie 1/100 - 1/1000 + 1/10000 ....... we can take the the positive term to some value up, then take the negative term to reduce it to some value down and our approximation will get better because the terms get smaller.
But this is done via choice, not algorithmic, so while I understand the idea, it feels incomplete and doesn't capture the full picture for me.
Is this also true for non-alternating series? Surely not, since we will only have either positive or negative terms.
>pic related
????
>>8198964
>But this is done via choice
Not a problem. We're doing math here and not CS. An existence argument is sufficient.
>Is this also true for non-alternating series?
It is true only for conditionally convergent series, i.e. those which converge but don't converge absolutely. This implies that there are infinitely many negative and infinitely many positive terms.
>>8198995
>Not a problem. We're doing math here and not CS. An existence argument is sufficient.
is this pretty much all I need to do show that every convergent alternating series can be manipulated to be equal to any value; QED?
>It is true only for conditionally convergent series
Then why can't I manipulate the sum of 1/(n^2) any way I can?
>>8198964
>alternating series
please do not use that word.
riemann showed that any convergent series that DOES NOT CONVERGE ABSOLUTELY can be rearranged to converge against anything we want
>>8199016
>Then why can't I manipulate the sum of 1/(n^2) any way I can?
Becuase it's abslutely convergent. All terms are positive.
>>8198995
>An existence argument is sufficient
Incorrect terminology but since you're being so informal anyways it's par for the course.
>>8199016
Axiom of choice is non-constructive, as the other anon pointed out. So you shouldn't expect to actually be able to use it. It's just a funny useless oddity where you can claim that there must exist a way to manipulate such series but you have probability 0 that such a manipulation can be defined or described in a finite amount of text or time.
>>8198964
>But this is done via choice, not algorithmic
Why do you think what you have described is not an algorithm? Take positive terms until the sum is greater than the target, then switch to negative terms until the sum is lower, and so on. That's definitely an algorithm.
>>8199050
An algorithm this simple won't work for every series. That is to say that one can easily construct a series for which this algorithm will fail.
>>8199016
>Then why can't I manipulate the sum of 1/(n^2) any way I can?
That is an absolutely convergent series, not a conditionally convergent one.
>>8199059
It should work for any conditionally convergent series.
>>8199061
Okay, so back to my original question, why can't I manipulate the series for 1/(n^2) any way i can?
In a conditionally convergent series, there seems to be some equivalence relation with the fact that it has infinitely many positive and negative terms so we can "exploit" this property and manipulate the result any way we want.
Surely this is not the case with a series that converge absolutely because there's no negative terms.
>>8198964
"I am guessing this have to do with the fact that we have infinite amount of positive and negative term, so it is possible to do arrangement such that we can use it to approximate the summation to any value without "exhaustion" in a sense. Since our series is an convergent alternating series, our approximation will get better."
yes
you can prove that it equals every number you want because eventually your'e doing 'infinity-infinity'
here's pi for example
source https://www.youtube.com/watch?v=-EtHF5ND3_s
>>8198964
Have you actually read the proof?
Generally if you don't understand a theorem studying the proof helps.
>>8199199
The proof for what theorem?
>>8198964
>Why is it this a problem
With a convergent sequence all subsequences converge to the same limit, with non-convergent sequences this isn't the case. And that's pretty much what you're doing when you rearrange a sum, you're extracting another "sub-sum". For convergent sums each sub-sum will converge to the same limit as the sum.
