∑n=1∞(sin⁡n+2)nn3nDoes it converge or diverge?

The values for which ${\displaystyle \sin \left(n\right)}$ is close to 1 (say in an interval ${\displaystyle [1-ϵ;1]}$) are some what regular:
${\displaystyle 1-ϵ\le \sin \left(n\right)}$ implies that there exists an integer k(n) such that ${\displaystyle n=2k\left(n\right)+\pi +\frac{\pi }{2}+a\left(n\right)}$ where ${\displaystyle \left|a\left(n\right)\right|\le \arccos (1-ϵ)}$. As ${\displaystyle ϵ\to 0,\arccos (1-ϵ)\sim \sqrt{2ϵ}}$, thus we can safely say that for ${\displaystyle ϵ}$ small enough, ${\displaystyle |n-2k\left(n\right)\pi -\frac{\pi }{2}|=\left|a\left(n\right)\right|\le 2\sqrt{ϵ}}$
If ${\displaystyle m>n}$ and since ${\displaystyle \sin \left(n\right)}$ and ${\displaystyle \sin \left(m\right)}$ are both in ${\displaystyle [1-ϵ;1]}$, then we have the inequality ${\displaystyle |(m-n)-2(k\left(m\right)-k\left(n\right))\pi |\le |m-2k\left(m\right)\pi -\frac{\pi }{2}|+|n-2k\left(n\right)\pi -\frac{\pi }{2}|\le 4\sqrt{ϵ}}$ where ${\displaystyle (k\left(m\right)-k\left(n\right))}$ is some integer k.
Since ${\displaystyle \pi }$ has a finite irrationality measure, we know that there is a finite real constant ${\displaystyle \mu >2}$ such that for any integers ${\displaystyle n,k}$ large enough, ${\displaystyle |n-k\pi |\ge k^{1-\mu }}$
By picking ${\displaystyle ϵ}$ small enough we can forget about the finite number of exceptions to the inequality, and we get ${\displaystyle 4\sqrt{ϵ}\ge {\left(2k\right)}^{1-\mu }}$. Thus
${\displaystyle (m-n)\ge 2k\pi -4\sqrt{ϵ}\ge \pi {\left(4\sqrt{ϵ}\right)}^{\frac{1}{1-\mu }}-4ϵ\ge A_{ϵ}=A{\sqrt{ϵ}}^{\frac{1}{1-\mu }}}$ 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 ${\displaystyle ϵ}$ gets smaller
We can get a lower bound for the first problematic term using the irrationality measure as well : from ${\displaystyle |n-2k\left(n\right)\pi -\frac{\pi }{2}|\le 2\sqrt{ϵ}}$, we get that for ${\displaystyle ϵ}$ small enough, ${\displaystyle {(4k+1)}^{1-\mu }\le |2n-(4k+1)\pi |\le 4\sqrt{ϵ}}$, and then ${\displaystyle n\ge B_{ϵ}=B{\sqrt{ϵ}}^{\frac{1}{1-\mu }}}$ for some constant B.
Therefore, there exists a constant C such that forall ${\displaystyle ϵ}$ small enough, the k-th integer n such that ${\displaystyle 1-ϵ\le \sin n}$ is greater than ${\displaystyle C_{ϵ}}$

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