Convergence/Divergence of infinite series ∑∞n=1(sinn+2)nn3n\sum_{n=1}^{\infty} \frac{(\sin n+2)^n}{n3^n}

The values for which $\sin(n)$ is close to $1$ (say in an interval $[1-\varepsilon ; 1]$) are somewhat regular :

$1 - \varepsilon \le \sin(n)$ implies that there exists an integer $k(n)$ such that
$n = 2k(n) \pi + \frac \pi 2 + a(n)$ where $|a(n)| \leq \arccos(1- \varepsilon)$.
As $\varepsilon \to 0$, $\arccos(1- \varepsilon) \sim \sqrt{2 \varepsilon}$, thus
we can safely say that for $\varepsilon$ small enough, $|n-2k(n) \pi - \frac{\pi}2| = |a(n)| \leq 2 \sqrt{ \varepsilon}$

If $m \gt n$ and $\sin(n)$ and $\sin(m)$ are both in $[1-\varepsilon ; 1]$,
then we have the inequality $|(m-n) - 2(k(m)-k(n)) \pi| \leq |m-2k(m)\pi - \frac{\pi}2| + |n-2k(n)\pi - \frac{\pi}2| \leq 4 \sqrt { \varepsilon}$ where $(k(m)-k(n))$ is some integer $k$.

Since $\pi$ has a finite irrationality measure, we know that there is a finite real constant $\mu \gt 2$ such that for any integers $n,k$ large enough,
$|n-k \pi| \ge k^{1- \mu}$.

By picking $\varepsilon$ small enough we can forget about the finite number of exceptions to the inequality, and we get $4\sqrt{\varepsilon} \ge (2k)^{1- \mu}$.
Thus $(m-n) \ge 2k\pi - 4\sqrt{\varepsilon} \ge \pi(4\sqrt{\varepsilon})^{\frac1{1- \mu}} - 4\sqrt{\varepsilon} \ge A_\varepsilon = A\sqrt{\varepsilon}^{\frac1{1- \mu}}$ for some constant $A$.

Therefore, we have a guarantee on the lengh of the gaps between equally problematic terms, and we know how this length grows as $\varepsilon$ gets smaller (as we look for more problematic terms)

We can get a lower bound for the first problematic term using the irrationality measure as well : from $|n-2k(n) \pi - \frac{\pi}2| \leq 2\sqrt {\varepsilon}$, we get that for $\varepsilon$ small enough, $(4k+1)^{1- \mu} \le |2n - (4k+1) \pi| \le 4\sqrt \varepsilon$, and then $n \ge B_\varepsilon = B\sqrt\varepsilon^{\frac1{1- \mu}}$ for some constant $B$.

Therefore, there exists a constant $C$ such that forall $\varepsilon$ small enough, the $k$-th integer $n$ such that $1-\varepsilon \le \sin n$ is greater than $C_\varepsilon k = C\sqrt\varepsilon^{\frac1{1- \mu}}k$

Since $\varepsilon < 1$ and $\frac 1 {1- \mu} < 0$, this bound $C_ \varepsilon$ grows when $\varepsilon$ gets smaller.
And furthermore, the speed of this growth is greater if we can pick a smaller (better) value for $\mu$ (though all that matters is that $\mu$ is finite)

Now let us give an upper bound on the contribution of the terms where $n$ is an integer such that $\sin (n) \in [1-2\varepsilon ; 1-\varepsilon]$

$$S_\varepsilon = \sum \frac{(2+\sin(n))^n}{n3^n} \le \sum_{k\ge 1} \frac{(1- \varepsilon/3)^{kC_{2\varepsilon}}}{kC_{2\varepsilon}} = \frac{- \log (1- (1- \varepsilon/3)^{C_{2\varepsilon}})}{C_{2\varepsilon}} \\ \le \frac{- \log (1- (1- C_{2\varepsilon} \varepsilon/3))}{C_{2\varepsilon}} = \frac{- \log (C_{2\varepsilon} \varepsilon/3))}{C_{2\varepsilon}}$$

$C_{2\varepsilon} = C \sqrt{2\varepsilon}^\frac 1 {1- \mu} = C' \varepsilon^\nu$ with $\nu = \frac 1 {2(1- \mu)} \in ] -1/2 ; 0[$, so :

$$S_\varepsilon \le - \frac{ \log (C'/3) + (1+ \nu) \log \varepsilon}{C'\varepsilon^\nu}$$

Finally, we have to check if the series $\sum S_{2^{-k}}$ converges or not :

$$\sum S_{2^{-k}} \le \sum - \frac { \log (C'/3) - k(1+ \nu) \log 2}{C' 2^{-k\nu}} = \sum (A+Bk)(2^ \nu)^k$$

Since $2^ \nu < 1$, the series converges.