Someone asked on Twitter:
Indeed this series diverges, just as the tweeter says: there’s a positive-density subset of n such that the summand exceeds 
More subtle: what about
This should still diverge. Argument: the probability that a real number x chosen uniformly from a large interval has 
What’s not so clear to me is: how does
grow with N?
Quomodocumque 
Typos? Looks like it should be “… is on order $\epsilon^{1/2}$, not $\epsilon$” and “… with density on order $\epsilon^{1/2}$.
Oops yeah fixed thanks
I think it grows like (log N)^{1/2}. As n grows, your window epsilon needs to shrink so that epsilon log n << 1 to keep the summand close to 1/n. So you end up with something like \Sum_{n=1}^N 1 / (n (log n)^{1/2}) ~ (log N)^{1/2}.
Jordan, can we steal this (and related) problems for the undergrad experimental math lab at PCMI this summer?
Please do! Though I have not thought about this enough to know how meaty it is. It feels like it has something to do with Diophantine approximation of pi by rationals, but I’m not sure what level of refinement is relevant.
Replacing
with 
, for random 
the expected size of 
is 
for an explicit constant 
: this follows from approximating 
and comparing with a Gaussian, so the typical growth is 
. It seems like you need only very mild equidistribution of 
. Are there are any bad irrational 
at all?
1/n^(2+cos(n)) seems pretty easy to deal with, since we don’t have any asymptotes to worry about; we can just treat cos(n) as being cos(x) for a random x in [0,2pi). This will give us some reasonable bounded random variable for each n, and then we can just sum that. Some empirical checks on W|A suggests that it’s a touch greater than 1/nlog(n), which certainly diverges.
It has been brought to my attention that I titled this post “A divergent sequence” when I meant “A divergent series” — well, the title will have to be understood to refer to the sequence of partial sums!
Don’t worry: the sequence diverges too! (in the sense that the terms of the original series do not converge to zero)