Convergence of $\sum_{n=1}^{\infty}\frac{(\frac{2}{3}+\frac{1}{3}\cdot \sin(n))^n}{n}$

Question

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.

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[1] Jonathan M. Borwein, David H. Bailey, and Roland Girgensohn. Experimentation in Mathematics: Computational Paths to Discovery. CRC Press, 2004.

Answer 1

I summarize below the main ideas of the paper [1] whose link is in my comment above:

1. 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.
2. 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.
3. 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.
4. 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