Re: sum R
- From: David C. Ullrich <dullrich@xxxxxxxxxxx>
- Date: Fri, 18 Apr 2008 06:50:55 -0500
On Thu, 17 Apr 2008 12:03:39 EDT, amy666 <tommy1729@xxxxxxxxxxx>
wrote:
On Apr 17, 8:40 am, David C. Ullrich
<dullr...@xxxxxxxxxxx> wrote:
On Wed, 16 Apr 2008 07:23:39 -0700 (PDT), TheQurqirish Dragon
wrote:
<qurqiri...@xxxxxxx> wrote:
On Apr 16, 7:59 am, amy666 <tommy1...@xxxxxxxxxxx>
interval.taking the sum over the reals in a real
internet.
i would like to read more about it on the
range (-1,1) the sum over the reals converges if the
in general for the interval (0,1) with bounded
integral over (0,1) is 0.
checked it carefully
That may be a necessary condition (I haven't
Consider the characteristicenough to be certain), but not sufficient.
matter, any f whichfunction on the rationals, for example. For that
whatever you want -is 0 almost everywhere can be used to make the sum
divergent. Although to beconvergent, conditionally convergent, or
of f needs to beconditionally convergent or divergent, this type
we're adding, thenon-0 on an infinite set.
No. Since there's no order given for the terms
sum cannot possible converge unless it convergesabsolutely.
In which case f vanishes except on a _countable_set.
David C. Ullrich
I assumed that the order of addition of terms was
that a_x is an
earlier term than a_y if x < y. (please excuse the
abuse of notation,
as I don't know if this is appropriate for a possibly
uncountable
sum). That's why I left the possibility of a
conditional convergence.
As for countable, in this context I believe my
stipulation that f is 0
almost everywhere covered that.
your assumption works fine in ZFC.
my sum is considered as a sort of average.
so +1 -1 +1 -1 ... = 0
PS. Yes, there _are_ summability methods that give 0
for that sum. There are also summability methods that
give 42 for the very same sum. Which is why you need to
be much more specific about what you mean.
my highest regards
tommy1729
David C. Ullrich
.
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