Timeline for Given four conditionally convergent series, is there a single sequence of naturals such that each corresponding subseries sums to $\pm\infty$?

7 events

when toggle format	what		by	license	comment
May 4, 2017 at 15:15	vote	accept	Will Brian
May 4, 2017 at 15:15	comment	added	Will Brian		I'm with Christian -- this is a lovely answer. I am both ashamed of how long I tried proving this for four series . . . and thankful that I can now stop. Thanks!
May 4, 2017 at 0:54	comment	added	fedja		@ChristianRemling BTW, Christian, once you like such problems, take a look at math.stackexchange.com/questions/2256846/… To my shame, I do not know how to handle 3 or more permutations...
May 4, 2017 at 0:47	comment	added	fedja		@ChristianRemling Thanks! I preserved the $\pm$ reformulation because it helps a bit when thinking about 3 series too. In fact, you have even more freedom: you can skip vectors as well due to the sign version of Levy-Steinitz. Once you realize that, the three series case becomes a relatively simple casework.
May 3, 2017 at 22:54	comment	added	Christian Remling		Elegant and simple (afterwards, that is) example! One can actually start reading at the end, and then the business with the signs is unnecessary: as you stated, the example now becomes $(-1)^n (1/k,1/k, 1/N_k, 0)$ etc., and your argument as stated can also be read as being about making selections within each block rather than imposing signs (what we look at is now called $\sum_{A_k}(-1)^n$ in this version).
May 3, 2017 at 19:50	history	edited	fedja	CC BY-SA 3.0	deleted 1 character in body
May 3, 2017 at 19:45	history	answered	fedja	CC BY-SA 3.0