5
$\begingroup$

The p-series convergence test is a classic and well-known result for sums of the form $\sum_{n=1}^{\infty}\frac{1}{n^p}$ for a real number $p$. It is known that $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges, but for every $\epsilon>0$, $\sum_{n=1}^{\infty}\frac{1}{n^{1+\epsilon}}$ converges.

It can be shown that series with terms asymptotically smaller than this will also converge, such as $$\sum_{n=2}^{\infty}\frac{1}{n\log^2n}\text{ and even }\sum_{n=2}^{\infty}\frac{1}{n\log^{1+\epsilon}n}\text{ for }\epsilon>0$$

I was introduced to a related series by a coworker of mine, which is the following: $$\sum_{n=1}^{\infty}\frac{1}{n^{1+\sin n}}$$ Supposedly, he was able to prove that this diverged. A natural generalization is to look at series of the form $$\sum_{n=1}^{\infty}\frac{1}{n^{c+\sin n}}$$ for some $c>0$. It is not hard to show that the series diverges when $c\leq0$ and converges when $c\geq2$. What I want is to find the smallest value of $c$ such that the series converges, or a tight lower bound. Formally, I want to find $$\inf\left\{c\,:\,\sum_{n=1}^{\infty}\frac{1}{n^{c+\sin n}}<\infty\right\}$$ Any progress on finding this number is appreciated. I would assume that it is greater than 1, but I haven't been able to prove much else.

$\endgroup$
12
  • 2
    $\begingroup$ In your scond paragraph do you mean "will also converge" or "will also diverge" ? $\endgroup$
    – Delta-u
    Commented Oct 12, 2018 at 14:11
  • $\begingroup$ I mean converge, also in addition to $\sum_{n=1}^{\infty}\frac{1}{n^{1+\epsilon}}$. The two I referenced can be shown from the integral test $\endgroup$
    – HackerBoss
    Commented Oct 12, 2018 at 14:29
  • $\begingroup$ @Delta-u Good catch though. Just realized I was dividing by zero. I have changed the lower bounds accordingly $\endgroup$
    – HackerBoss
    Commented Oct 12, 2018 at 14:44
  • $\begingroup$ Anyway on different note I think that when $c<2$ the series diverges, because $\sin n \in[c-1,1]$ infinitely often and the probability that $\sin n \in [c-1,1]$ is non-zero (though I can't prove this statement). $\endgroup$
    – kingW3
    Commented Oct 12, 2018 at 14:57
  • $\begingroup$ @kingW3 I have just corrected that part of the question again. They actually do converge now $\endgroup$
    – HackerBoss
    Commented Oct 12, 2018 at 14:58

1 Answer 1

1
$\begingroup$

The series clearly diverges for any $c<2$. To see this at a glance suppose $c = 2-\epsilon$. The idea is to look at the unit circle and note the proportion of angles for which the number of terms with $c+\sin(n)\leq1$ is finite. $$\begin{equation} \sum_{n=0}^N \frac{1}{n^{c + \sin n}} \geq \sum_{n \in S_N} \frac{1}{n^{2 - \epsilon + \sin n}} \end{equation}$$ where $S_N = \Big\{n: \sin n \leq -1 + \epsilon \ \ \text{and} \ \ 0 \leq n \leq N\Big\}$ and $0<\delta<\epsilon$. Then $|S_N|/N$ goes to $\cos^{-1}(1-\epsilon)$ as $N$ goes to infinity by equidistribution of $n\mod 2\pi$ and the RHS becomes of the same order as $\sum 1/n$.

Edit: $\delta$ was unnecessary

$\endgroup$
3
  • 1
    $\begingroup$ Over the real numbers, that would be true. The trick here is that we are looking only at integers. Can you prove that $|S_N|/N\rightarrow(\epsilon - \delta)/2\pi$ as $N\rightarrow\infty$ looking at only integral values of $n$? $\endgroup$
    – HackerBoss
    Commented Oct 12, 2018 at 15:10
  • $\begingroup$ Last paragraph is unnecessary. Just refer to equidistribution of $n$ mod $2\pi$. en.wikipedia.org/wiki/Equidistributed_sequence $\endgroup$ Commented Oct 12, 2018 at 15:35
  • $\begingroup$ The limit of $|S_N|/N$ is $\frac 1{\pi} \cos^{-1} (1-\epsilon+\delta)$. But this doesn't affect the validity of this argument. $\endgroup$ Commented Oct 12, 2018 at 15:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .