# On convergence of sums of the form $\sum_{n=1}^{\infty}\frac{1}{n^{1+f(n)}}$

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.

• In your scond paragraph do you mean "will also converge" or "will also diverge" ? – Delta-u Oct 12 '18 at 14:11
• 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 – HackerBoss Oct 12 '18 at 14:29
• @Delta-u Good catch though. Just realized I was dividing by zero. I have changed the lower bounds accordingly – HackerBoss Oct 12 '18 at 14:44
• 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). – kingW3 Oct 12 '18 at 14:57
• @kingW3 I have just corrected that part of the question again. They actually do converge now – HackerBoss Oct 12 '18 at 14:58

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
• 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 $n$? – HackerBoss Oct 12 '18 at 15:10
• Last paragraph is unnecessary. Just refer to equidistribution of $n$ mod $2\pi$. en.wikipedia.org/wiki/Equidistributed_sequence – Sungjin Kim Oct 12 '18 at 15:35
• 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. – Sungjin Kim Oct 12 '18 at 15:53