Does the sequence $\{\sin(en)\}$ converge or diverge?

Question

Is it known if $\{\sin(en)\}$ converges or diverges?

Also, I have a more general question. For almost every rational $r$, I think we can say that $\{\sin(rn)\}$ diverges. Does that statement hold if we only assume $r$ is a real number?

Answer 1

The sequence $(\sin rn)_{n\in\mathbb N}$ converges if and only if $r$ is a multiple of $\pi$ (in which case it is the constant sequence $0,0,0,\ldots$).

If $r$ is any rational-but-not-integral multiple of $\pi$, then the sequence is periodic but not constant, and therefore doesn't converge.

If $r/\pi$ is irrational, then values in the sequence are dense in $(-1,1)$ -- in particular there are infinitely many elements in $(-1,-\frac12)$ and also infinitely many elements in $(\frac12,1)$, so it cannot converge in this case either.

Answer 2

It does not convergence for all $e$ and therfore diverges. We can easily see this if we pick $e = \frac{\pi}{2}$. For $n \in \mathbb{N}$ we now see that $\sin(\frac{\pi}{2} n) = 0$ if $n$ is even. While if $n$ is odd, this is either $-1$ or $1$. This is not convergent and thus divergent.

Answer 3

Henning Makholm's answer is perfectly fine, but if you like another approach, it is straightforward to check that $\{\sin(en)\}$ cannot be a Cauchy sequence, since: $$ \limsup_{n\in\mathbb{N}}\left| \sin((n+1)e)-\sin(ne) \right|= 2\sin\frac{e}{2}\cdot\limsup_{n\in\mathbb{N}}\left|\cos\left((2n+1)\frac{e}{2}\right)\right|\color{red}{>}1. $$

Answer 4

You can use

$$ \sum_{k=1}^n \sin(kx) = \tfrac{1}{2} \sin(nx) + \tfrac{1}{2} \sin(x) \frac{ 1 - \cos(nx) }{ 1 - \cos(x) } $$

When you put in $x=e$, you get

$$ \sum_{k=1}^n \sin(ke) = \tfrac{1}{2} \sin(ne) + \tfrac{1}{2} \sin(e) \frac{ 1 - \cos(ne) }{ 1 - \cos(e) }, $$

so

$$ \lim_{n \rightarrow \infty} \sum_{k=1}^n \sin(ke) = \lim_{n \rightarrow \infty} \tfrac{1}{2} \left\{ \sin(ne) + \sin(e) \frac{ 1 - \cos(ne) }{ 1 - \cos(e) } \right\}, $$ which is undefined, i.e. it does not converge.

Answer 5

It's worth mentioning that if we consider arbitrary $s$ then the series

$$ \sum_{n=0}^{\infty} \sin(sn) $$

Can usually still be assigned a formal divergent sum in a way similar to how $1 - 1 + 1 - 1 ... = \frac{1}{2}$

Recall that $$\sin(n) = \frac{e^{in} - e^{-in}}{2i} $$

So the series

$$ \sum_{n=0}^{\infty} \sin(sn)x^n = \frac{1}{2i} \left(\sum_{n=0}^{\infty} e^{isn}x^n - \sum_{n=0}^{\infty} e^{-isn}x^n \right) = $$

$$ \frac{1}{2i} \left( \frac{1}{1-e^{is}x} - \frac{1}{1-e^{-is}x} \right) $$

Now evaluating at $x=1$ we find that

$$ \sum_{n=0}^{\infty} \sin(sn) = \frac{1}{2i} \left( \frac{1}{1-e^{is}} - \frac{1}{1-e^{-is}} \right) $$

For that particular problem then we have

$$ \sum_{n=0}^{\infty} \sin(en) = \frac{1}{2i} \left( \frac{1}{1-e^{ei}} - \frac{1}{1-e^{-ei}} \right) $$