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 $0$. $\square$

Answer 2

By request of bounty, I'll present a more detailed and clear proof to this question. Please notice that this exact question, both about the sequence and about the summation, have a large number of duplicates. In chronological order: here and here and here and here and here and here and here and here and here. This post is only the second oldest version I could find.

I will prove that the sequence diverges, which immediately proves that the sum diverges too.

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We prove that the sequence $\{\sin(n)^n\}_{n\in\mathbb{N}}$ has a limit supremum of $1$. The standard technique is to use a variation on Dirichlet's approximation theorem. In particular, it's possible to find infinitely many odd integers $p_n, q_n$ such that the following holds. $$\left|\frac{\pi}{2}-\frac{p_n}{q_n}\right| < \frac{1}{q_n^2}$$

We may assume $q_n\to \infty$. The above can be rewritten using big-O notation, to simply say $\frac{\pi}{2} = \frac{p_n}{q_n} + \mathcal{O}(q_n^{-2})$, which also implies $q_n\frac{\pi}{2}-p_n = \mathcal{O}(1/q_n)$. Using these facts, we estimate $|\sin(p_n)|$ like so. $$\begin{align} |\sin(p_n)| &= \left|\cos\left(q_n\frac{\pi}{2}-p_n\right)\right| \\ &= \cos(\mathcal{O}(1/q_n)) \\ &= 1+\mathcal{O}(1/q_n^2) \\ &= 1+\mathcal{O}(1/p_n^2) \end{align}$$

The first equality holds since $q_n$ is odd, the second follows from previous observations, the third is the small-angle approximation for cosine, and the last follows since $p_n\sim q_n\pi/2$. Anyway, now we estimate $|\sin(p)|^p$. $$\begin{align} |\sin(p_n)|^{p_n} &=(1+\mathcal{O}(1/p_n^2))^{p_n} \\ &=\exp(p_n\ln(1+\mathcal{O}(p_n^{-2})) \\ &= \exp(p_n\cdot\mathcal{O}(1/p_n^2)) \\ &= \exp(\mathcal{O}(1/p_n) \\ &= 1 + \mathcal{O}(1/p_n) \end{align}$$

By limiting $n\to\infty$, we conclude that $|\sin(p_n)|^{p_n}\to 1$. This shows that $|\sin(n)|^n$ has a limit supremum of $1$. It's trivial to find a subsequence where $\sin(n)^n$ converges to zero, so the sequence cannot converge. A clever variation on the above technique also proves the stronger result that $\{\sin(n)^n\}_{n\in\mathbb{N}}$ is dense on the interval $[-1,1]$.