A divergent sequence

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 n^{-1/2}.

More subtle: what about

\sum_{n=1}^{\infty} \frac{1}{n^{2 + \cos n}}?

This should still diverge. Argument: the probability that a real number x chosen uniformly from a large interval has \cos x < -1+\epsilon is on order \epsilon^{1/2}, not \epsilon; so there will be a subset of integers with density on order \epsilon^{1/2} where the summand exceeds n^{-1-\epsilon}, and summing over those integers along should give a sum on order \epsilon^{-1/2}, which can be made as large as you like by bringing \epsilon close to 0.

What’s not so clear to me is: how does

\sum_{n=1}^{N} \frac{1}{n^{2 + \cos n}}

grow with N?

Tagged , ,

9 thoughts on “A divergent sequence

  1. Daniel Weissman says:

    Typos? Looks like it should be “… is on order $\epsilon^{1/2}$, not $\epsilon$” and “… with density on order $\epsilon^{1/2}$.

  2. JSE says:

    Oops yeah fixed thanks

  3. Daniel Weissman says:

    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}.

  4. Jayadev Athreya says:

    Jordan, can we steal this (and related) problems for the undergrad experimental math lab at PCMI this summer?

  5. JSE says:

    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.

  6. Sean Eberhard says:

    Replacing \cos n with \cos(2 \pi \alpha n), for random \alpha the expected size of n^{-2+\cos(2 \pi \alpha n)} is c n / \sqrt{\log n} for an explicit constant c: this follows from approximating \cos x \approx 1 - x^2/2 and comparing with a Gaussian, so the typical growth is c'\sqrt{\log N}. It seems like you need only very mild equidistribution of \alpha. Are there are any bad irrational \alpha at all?

  7. 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.

  8. JSE says:

    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!

  9. Sean Eberhard says:

    Don’t worry: the sequence diverges too! (in the sense that the terms of the original series do not converge to zero)

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: