2022-02-23

$\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(\mathrm{sin}n+2\right)}^{n}}{n{3}^{n}}$
Does it converge or diverge?

poskakatiqzk

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

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