This is inspired by this Math.SE question, for $a=1$.
Borwein, Bailey, and Girgensohn pose in their book ([1,Problem 35]) as an open problem the convergence of the series $$\sum_{n=1}^\infty \frac{(2+\sin n)^n}{3^n n}$$ This was solved in 2005 by Ravi B. Boppana, who put a preprint on arXiv with the solution in 2020 [2]. The solution is based on dividing this series in two, to handle separately the contribution of (1) the "tame" $n$'s for which $\sin n$ is bounded away from $1$ (by a gap at least $1/\sqrt{n}$) and (2) the "wild" ones, which can be very arbitrarily close to $1$, but that the proof shows are sparse enough that the sum of their reciprocals converges. Crucially, the argument for this second case relies on a bound on the irrationality measure $\mu(\pi)$ of $\pi$: $\mu(\pi) \leq 20$ suffices.
If I am not mistaken, even assuming $\mu(\pi)=2$ (the best possible) this proof technique, balancing the parameters, would break for $a=2/3$: $$\sum_{n=1}^\infty \frac{(2+\sin n)^n}{3^n n^{2/3}}$$ (maybe worth checking, but this is what my back-of-the-envelope calculations led to). Which leads me to my question:
Using any available techniques, what can we say about the convergence of $$\sum_{n=1}^\infty \frac{(2+\sin n)^n}{3^n n^a}$$ as a function of $a\in(0,1)$?
And does the implication go the other way -- that is, does proving convergence (or divergence) for any particular $a\in(0,1)$ imply an upper or lower bound on $\mu(\pi)$?
[1] Jonathan M. Borwein, David H. Bailey, and Roland Girgensohn. Experimentation in Mathematics: Computational Paths to Discovery. CRC Press, 2004.
[2] Ravi B. Boppana. Convergence of a sinusoidal infinite series from Borwein, Bailey, and Girgensohn. 2020. arXiv:2007.11017
