Nyah Conrad

2022-02-23

Does it converge or diverge?

poskakatiqzk

Beginner2022-02-24Added 8 answers

The values for which $\mathrm{sin}\left(n\right)$ is close to 1 (say in an interval $[1-\u03f5;\text{}1]$ ) are some what regular:

$1-\u03f5\le \mathrm{sin}\left(n\right)$ implies that there exists an integer k(n) such that $n=2k\left(n\right)+\pi +\frac{\pi}{2}+a\left(n\right)$ where $\left|a\left(n\right)\right|\le \mathrm{arccos}(1-\u03f5)$ . As $\u03f5\to 0,\text{}\mathrm{arccos}(1-\u03f5)\sim \sqrt{2\u03f5}$ , thus we can safely say that for $\u03f5$ small enough, $|n-2k\left(n\right)\pi -\frac{\pi}{2}|=\left|a\left(n\right)\right|\le 2\sqrt{\u03f5}$

If$m>n$ and since $\mathrm{sin}\left(n\right)$ and $\mathrm{sin}\left(m\right)$ are both in $[1-\u03f5;1]$ , then we have the inequality $|(m-n)-2(k\left(m\right)-k\left(n\right))\pi |\le |m-2k\left(m\right)\pi -\frac{\pi}{2}|+|n-2k\left(n\right)\pi -\frac{\pi}{2}|\le 4\sqrt{\u03f5}$ where $(k\left(m\right)-k\left(n\right))$ 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$\u03f5$ small enough we can forget about the finite number of exceptions to the inequality, and we get $4\sqrt{\u03f5}\ge {\left(2k\right)}^{1-\mu}$ . Thus

$(m-n)\ge 2k\pi -4\sqrt{\u03f5}\ge \pi {\left(4\sqrt{\u03f5}\right)}^{\frac{1}{1-\mu}}-4\u03f5\ge {A}_{\u03f5}=A{\sqrt{\u03f5}}^{\frac{1}{1-\mu}}$ 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$\u03f5$ gets smaller

We can get a lower bound for the first problematic term using the irrationality measure as well : from$|n-2k\left(n\right)\pi -\frac{\pi}{2}|\le 2\sqrt{\u03f5}$ , we get that for $\u03f5$ small enough, $(4k+1)}^{1-\mu}\le |2n-(4k+1)\pi |\le 4\sqrt{\u03f5$ , and then $n\ge {B}_{\u03f5}=B{\sqrt{\u03f5}}^{\frac{1}{1-\mu}}$ for some constant B.

Therefore, there exists a constant C such that forall$\u03f5$ small enough, the k-th integer n such that $1-\u03f5\le \mathrm{sin}n$ is greater than $C}_{\u03f5$

If

Since

By picking

Therefore, we have a guarantee on the lengh of the gaps between equally problematic terms, and we know how this length grows as

We can get a lower bound for the first problematic term using the irrationality measure as well : from

Therefore, there exists a constant C such that forall

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