Does the sequence $\{\sin^n(n)\}$ converge?

Question

Does the sequence $\{\sin^n(n)\}$ converge?

Does the series $\sum\limits_{n=1}^\infty \sin^n(n)$ converge?

Answer 1

claim. The series $\sum\limits_{n=1}^\infty \sin^n(n)$ diverges.

Lemma. For all number $x$ irrational there exist a rational sequence $\{\frac{p_n}{q_n}\}$ where $\{q_n\}$ is odd such that $$ \left\vert x-\frac{p_n}{q_n}\right\vert<\frac{1}{q_n^2} $$

Proof.

Define $x_n=\frac{1}{x_{n-1}-\lfloor x_{n-1}\rfloor}$. Let $a_0=\lfloor x\rfloor$ and $a_n=\lfloor x_n\rfloor$ for $n\in \mathbb{N}$.

Let $R_n=a_0+\frac{1\vert}{\vert a_1}+\cdots+\frac{1\vert}{a_n\vert}$ wich denotes continued fraction

We have $x_{n+1}=a_{n+1}+\frac{1}{x_{n+2}}>a_{n+1}$ then we obtain that $$(q_n x_{n+1}+q_{n-1})q_n>(q_n a_{n+1}+q_{n-1})q_n=q_{n+1}q_n.$$

Then using an another result : $$\left\vert x-R_n\right\vert <\frac{1}{q_nq_{n+1}}$$ As $q_{n+1}=q_n a_{n+1}+q_{n-1}>q_na_{n+1}>q_n$ One can prove that the sequence $(q_n)$ contains an infinite odd number.

Indeed we have $$p_{k-1}q_k-q_{k-1}p_k=(-1)^k$$ for $k=1,2,\cdots,n$ and use Bézout's theorem.

For a general result look at : Irrationality Measure

Using the lemma for $x=\frac{\pi}{2}$ we have $\vert \sin(p_n)\vert=\vert \cos (\frac{\pi}{2}q_n -p_n)\vert>\cos (\frac{1}{q_n})>1-\frac{1}{2q_n^2}$ then the sequence $\{(\sin(p_n)^{p_n}\}$ does not converges to $0$. $\square$