For fixed p in (0,1], consider the formal series S(p) = Sum_{1 <= n < oo} e_n / n^p where each e_n = +-1 independently at random. Let f(p) := Prob(S(p) converges) Puzzle: Find f(p). —Dan
On Sep 10, 2015, at 11:13 AM, rwg <rwg@sdf.org> wrote:
On 2015-09-10 10:44, Dan Asimov wrote:
That was by far the most interesting thing I learned in 18.03. —Dan
On Sep 10, 2015, at 9:23 AM, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote: By suitable rearrangement, a conditionally convergent series can sum to any arbitrarily chosen value. Furthermore, a rearrangement exists such that the successive sums have any arbitrarily chosen limsup and liminf.
In fact, when limsup < liminf, it's *extremely* interesting. --rwg
Just rearrange that the set of terms of the series be the empty set.