Convergence by changing sign of terms
Question:
For two sequences both , can we always choose signs so that both and converge? What about three or more sequences?
Thoughts:
First consider only one sequence . Partial sums are monotone increasing, so either converge to finite limit or diverge to + infinity.
If converges, then done with . If diverges, then we can change the signs to make converge to any limit , by making the partial sums oscillate around L. (similar argument to rearrangement theorem for conditionally convergent series, but here don't reorder terms, only change signs).
So for one sequence the result is clearly true, and in fact there are uncountably many choices of the signs (since at least one choice for each limit L) for each divergent sequence .
For two sequences I have tried some ideas but I cannot prove it or construct a counterexample.
For two sequences both , can we always choose signs so that both and converge? What about three or more sequences?
Thoughts:
First consider only one sequence . Partial sums are monotone increasing, so either converge to finite limit or diverge to + infinity.
If converges, then done with . If diverges, then we can change the signs to make converge to any limit , by making the partial sums oscillate around L. (similar argument to rearrangement theorem for conditionally convergent series, but here don't reorder terms, only change signs).
So for one sequence the result is clearly true, and in fact there are uncountably many choices of the signs (since at least one choice for each limit L) for each divergent sequence .
For two sequences I have tried some ideas but I cannot prove it or construct a counterexample.
I have a new idea:
If we can first prove that for any complex sequence we can choose signs so that converges, then we can take real and imaginary parts, and then we get two real sequences and the sums and must converge. This would prove the question true, but still need to prove this complex case now!
If we can first prove that for any complex sequence we can choose signs so that converges, then we can take real and imaginary parts, and then we get two real sequences and the sums and must converge. This would prove the question true, but still need to prove this complex case now!
Original post by liewchuan
I have a new idea:
If we can first prove that for any complex sequence we can choose signs so that converges, then we can take real and imaginary parts, and then we get two real sequences and the sums and must converge. This would prove the question true, but still need to prove this complex case now!
If we can first prove that for any complex sequence we can choose signs so that converges, then we can take real and imaginary parts, and then we get two real sequences and the sums and must converge. This would prove the question true, but still need to prove this complex case now!
You might find some mileage in this http://forums.xkcd.com/viewtopic.php?f=3&t=10612, but I've not read it all through.
(edited 13 years ago)
Original post by ghostwalker
You might find some mileage in this http://forums.xkcd.com/viewtopic.php?f=3&t=10612, but I've not read it all through.
I found this proof :
http://books.google.co.uk/books?id=ButlynVk25MC&pg=PA151
http://books.google.co.uk/books?id=ButlynVk25MC&pg=PA151
Original post by DFranklin
Good find, and it has a complete solution when you do get to the end.
Cheers. I rely on you expertise as to it having a complete solution; I'll have to read it through carefully when I'm sober! Indexing takes a lot of thinking about; well, for me.
(edited 13 years ago)
Original post by ghostwalker
Cheers. I rely on you expertise as to it having a complete solution; I'll have to read it through carefully when I'm sober! Index sets take a lot of thinking about; well, for me.
Just to put it out there, Bollobas' room is in my staircase.
Original post by liewchuan
I'm not sure why the "optimal" value of the bound is in Bollobas? In the other forum link, the argument suggests it is . Probably it doesn't matter, but it's making me doubt the validity of the forum proof.
I think I'm going to see if I can find a counterexample to the sqrt(2) case.
OK, Bela is right (no surprise).
As I understand it, the "gotcha" is that life is easy if you replace z1,z2,z3,... by z1+z2, z3... (or z1-z2, z3, ...) but if you instead replace with z1+z3, z2, ..., then one of the sums you need to bound is , and you can't control whether it's + or -, because that's being specified by the induction hypothesis.
So the question is, "what's the worst that can happen?". Obviously for a worst case, we want z1 + z2 as big as possible (where WLOG we've taken the + sign), which would mean z1 and z2 being in the same direction (or nearly so). But that can't happen, because if z1 + z2 is large, then z1-z2 must be small, and we can use the "life is easy" case. It turns out (think equilateral triangles) that if we insist that the smaller of |z1+z2|, |z1-z2| is greater than 1, then the larger value can't be bigger than sqrt(3).
Similar arguments apply in the case where you replace with z2+z3.
I have to say, this key point was not obvious to me from the solution in Bela's book.
As I understand it, the "gotcha" is that life is easy if you replace z1,z2,z3,... by z1+z2, z3... (or z1-z2, z3, ...) but if you instead replace with z1+z3, z2, ..., then one of the sums you need to bound is , and you can't control whether it's + or -, because that's being specified by the induction hypothesis.
So the question is, "what's the worst that can happen?". Obviously for a worst case, we want z1 + z2 as big as possible (where WLOG we've taken the + sign), which would mean z1 and z2 being in the same direction (or nearly so). But that can't happen, because if z1 + z2 is large, then z1-z2 must be small, and we can use the "life is easy" case. It turns out (think equilateral triangles) that if we insist that the smaller of |z1+z2|, |z1-z2| is greater than 1, then the larger value can't be bigger than sqrt(3).
Similar arguments apply in the case where you replace with z2+z3.
I have to say, this key point was not obvious to me from the solution in Bela's book.
Original post by liewchuan
Thanks, probably he skipped the optimality because as long as you have such a constant C for n=1,2, you can make the inductive step and keep the same C from that point on (it doesn't matter what C actually is). But also, he clearly wanted to say that he knew the optimal value.
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