Convergence of $\sum_{n=1}^{\infty}\frac{(\frac{2}{3}+\frac{1}{3}\cdot \sin(n))^n}{n}$ My friend asked me this problem:
Source : Problem 35 from a 2004 book by Borwein, Bailey, and Girgensohn [1]

Determine whether the series $$\sum_{n=1}^{\infty}\frac{(\frac{2}{3}+\frac{1}{3}\cdot \sin(n))^n}{n}$$ converges.

Firstly I want to use the root test. But later I found that it just didn't work because $2/3+1/3\cdot \sin(n)$ can't be smaller than any given constant that is smaller than 1. Then I have no other ways to deal with it.

[1] Jonathan M. Borwein, David H. Bailey, and Roland Girgensohn. Experimentation in Mathematics: Computational Paths to Discovery. CRC Press, 2004.
 A: I summarize below the main ideas of the paper [1] whose link is in my comment above:

*

*Group the terms in the sum into "tame" and "wild" terms. The  tame  are defined as intergers that obey:
$$
\bigg|n-\frac{\pi}2-2\pi a\bigg|\ge\frac1{n^{1/4}}
$$
($a$ integer) meaning they are "far enough" from making the sine equal to $1$. Wilds are the non-tame integers.

*Using the following  theorem about how close $\pi$ is to rational numbers:
For every integers $p,q$ such that $|q|>1$:
$$
\bigg|\pi-\frac pq\bigg|>\frac1{|q|^{20}},
$$
they show that the  wild numbers  $W_k$ obey
$$
W_k\ge\frac12 k^{77/76}
$$
meaning they are pretty scarce.

*By using simple small angle expansion of the sine function they show that the sum over the  tame numbers is less than the sum of $e^{-\sqrt n}$ and therefore converges.

*Because of their scarcity the sum over the wild  numbers $W_k$ is less than or equal to twice the sum over $\frac1{k^{77/76}}$ and therefore also converges. Thus the whole sum converges.


[1] Convergence of a sinusoidal infinite series from Borwein, Bailey, and Girgensohn, Ravi B. Boppana (2020). arXiv:2007.11017


