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

Question

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.

Answer 1

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