Graph of the infinite sum $g(x)=\sum_{k=1}^\infty \sin^k(kx)$

Question

For which real numbers $x$ does the infinite sum $\sin(x)+\sin^2(2x)+\sin^3(3x)+ \cdots+\sin^n(nx)+\cdots$ converge?

How does its graph look?

Answer 1

Hint: Let $x=r \pi$.

If $r$ is irrational, then the set $nx -2k\pi$ is dense in $\mathbb{R}$. It is easy to get from here that $\lim_n \sin(nx)$ does not exist....

If $r=\frac{m}{k}$ with $gcd(m,k)=1$ and $\lim_n \sin(n\frac{m}{k} \pi)=0$ what can you say about $k$?

Answer 2

Let $x=r\pi$ and consider the following cases:

1. $r=\frac{a}{2b}$: For infinitely many $n$ we have $n=(2m+1)b$ thus $\sin^n(nx)=\sin^n\left(\frac{(2m+1)a}{2}\pi\right)=\pm 1$ so the sequence does not go to $0$ hence the series does not converge.
2. $r=\frac{a}{2b+1}$: For any $m$ we have $\left|nx-\frac{2m+1}{2}\pi\right|=\pi\left|\frac{2na-(2m+1)(2b+1)}{4b+2}\right|\geq \frac{\pi}{4b+2}$ and so $|\sin(nx)|$ is bounded above by $\lambda=\sin\left(\frac{2b}{2b+1}\pi\right)<1$, hence we have $$\sum\limits_{n=1}^\infty |\sin^n(nx)|<\sum\limits_{n=1}^\infty \lambda^n=\frac{\lambda}{1-\lambda}$$ so the series converges absolutely.
3. $r$ is irrational: This is more difficult. Let $[r]$ denote the fractional part of $r$. The key is to examine terms which are nearly half-integers. Let $N=\lfloor1/[r]\rfloor$. Consider the points $[r],[2r],\ldots,[nr]$ in $[0,1)$ ordered as real numbers. It is a theorem that there exist $\alpha>\beta>\gamma$ such that the distance between any two consecutive points is one of these three, and that the number of pairs with distance equal to each of these values differs by at most $N$. In particular, there are at least $\frac{n-2N}{3}$ pairs a distance $\alpha$ apart, so $\alpha<\frac{3}{n-2N}$. Thus by the pigeonhole principle, we have some $m\leq n$ such that $\left|[mr]-\frac12\right|<\frac{3}{n-2N}$. For such $m$ we have $$|\sin(mx)|>\left|\sin\left(\frac{(2k+1)\pi}{2}-\frac{3}{n-2N}\right)\right|=\cos\left(\frac{3}{n-2N}\right)>1-\frac{9}{2(n-2N)^2}$$ thus we get $$|\sin^m(mx)|>\left(1-\frac{9}{2(n-2N)^2}\right)^m>1-\frac{9m}{2(n-2N)^2}\geq 1-\frac{9n}{2(n-2N)^2}\to 1$$ and so the sequence does not converge to $0$, hence the series does not converge.

Since this does series not converge except on a set of measure $0$, I'm not sure how to meaningfully describe its graph. However, if you want to compute its value for $r=\frac{a}{2b+1}$ you can break the series up into sums over equivalence classes $\bmod 2b+1$ using absolute convergence, which gives $$\begin{align} \sum_{n=1}^\infty \sin^n(nx)&=\sum_{i=1}^{2b+1}\sum_{j=0}^\infty(-1)^j\sin^{j(2b+1)+i}\left(\frac{a+i}{2b+1}\pi\right)\\ &=\sum_{i=1}^{2b+1}\sin^i\left(\frac{a+i}{2b+1}\pi\right)\sum_{j=0}^\infty(-1)^j\sin^{j(2b+1)}\left(\frac{a+i}{2b+1}\pi\right)\\ &=\sum_{i=1}^{2b+1}\sin^i\left(\frac{a+i}{2b+1}\pi\right)\frac{1}{1+\sin^{(2b+1)}\left(\frac{a+i}{2b+1}\pi\right)}\\ \end{align}$$ which, while not the prettiest, is at least a finite sum.