Yes, any divergent series whose terms approach 0. Say we're choosing the sign of each term. Then all possible sign choices amount to the countable product K (cartesian power) of {-,+}, which has the size of 2^aleph_0 = the continuum. The product K naturally has the topology of the Cantor set. So for each point of the Cantor set K — i.e., for each choice of signs S : Z+ —> {-,+} — this S gives rise to the real number Sum_{n in Z+} S(n)/n. So assuming the probability of convergence is 1 (which it is), we get a map K —> R from the Cantor set onto the real numbers. This should by rights be continuous, so it's a funny map — a continuous surjection from the totally disconnected Cantor set onto the real numbers. But basic theorems in topology rule this out (the image must be compact). Therefore, mathematics is inconsistent. Another related thing is to look at the distribution of values obtained when choosing signs randomly. Or for more fun, use points on the unit circle S^1 in the complex plane, chosen independently at random from S^1, as the coefficient for each term of the harmonic series. What does the distribution of sums in the plane look like? —Dan Cris Moore wrote: ----- This applies to any divergent series, right? And the fun thing is that any C has a series with any initial subsequence. I suppose this means any C has countably many sequences, although it’s not clear that taking a finite initial subsequence and then turning greedy covers all of them…
On Dec 10, 2018, at 3:43 PM, Dan Asimov <dasimov@earthlink.net> wrote:
There is the greedy-harmonic-series-with-signs method: ...