Convergence and divergence depending on whether $n$ is odd or even

Question

It is part of one problem I am working on: I want to prove the following conjecture ($x\ne q\pi$ where $q\in\Bbb Q$)

$$\sum_{n=1}^{\infty}\frac{\sin^{2k-1}(nx)}{n^{\alpha}}\quad\text{converges if}\quad0<\alpha\le1,k\in\Bbb N$$ $$\sum_{n=1}^{\infty}\frac{\sin^{2k}(nx)}{n^{\alpha}}\quad\text{diverges if}\quad0<\alpha\le1,k\in\Bbb N$$

So far I have tested cases for $\sin^{1,2,3,4}$ and the results agree with the conjecture. I believe it holds for all natural numbers. I tried using induction but I failed. Could you help me or improve my method? Best regards.

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My failed induction method, for the $\sin^{2k}$ cases, was as follows:

Hypothesis
$$\sin^{2k}\theta=\xi_k+T_k(\theta)$$ where $\xi_k\in\Bbb R^+$ (the positivity of $\xi$ is needed in the context of the bigger problem, though it is not necessary in the conjecture), $T_k(\theta)$ is a linear combination of "first-order" trignometric terms: $\sin\theta,\sin 2\theta,\sin 3\theta,\cdots$ or $\cos\theta,\cos2\theta,\cos3\theta\cdots$ but not $\sin^2\theta,\sin^3 2\theta,\cos^45\theta$ etc. If this hypothesis holds, then by Abel-Dirichlet criterion it is easy to prove the second part of the conjecture.

Proof by induction
Initial case $k=1$ is trivial.
If for any given $k$ the hypothesis stands, then for $(k+1)$
$$ \begin{align*} \sin^{2k+2}\theta=&\xi_k\sin^2\theta+\sin^2\theta T_k(\theta)\\ =&\xi_k\left(\frac{1-\cos 2\theta}{2}\right)+\left(\frac{1-\cos 2\theta}{2}\right)T_k(\theta) \\ =&\frac12\xi_k-\frac12\xi_k\cos2\theta+\frac12T_k(\theta)-\frac12\cos(2\theta)T_k(\theta) \end{align*} $$ Now the hard part lies in the last term of RHS: it definitely contains terms like $\cos2\theta\sin m\theta$ or $\cos2\theta\cos m\theta$, to reduce them to the "first order", it must yield some constant terms, which might threaten the existence of a positive $\xi_{k+1}$.

I don't know how to proceed.

Answer 1

$$ \begin{align} \sum_{n=1}^\infty\frac{\sin^m(nx)}{n^\alpha} &=\sum_{n=1}^\infty\sum_{j=0}^m\frac{(-1)^j}{(2i)^m}\binom{m}{j}\frac{e^{i(m-2j)nx}}{n^\alpha}\tag{1}\\ &=\sum_{j=0}^m\frac{(-1)^j}{(2i)^m}\binom{m}{j}\sum_{n=1}^\infty\frac{e^{i(m-2j)nx}}{n^\alpha}\tag{2} \end{align} $$

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Case $\boldsymbol{m}$ is odd

Note that if $(m-2j)x\not\in2\pi\mathbb{Z}$, then $$ \begin{align} \left|\sum_{n=1}^N e^{i(m-2j)nx}\right| &=\left|\frac{e^{i(m-2j)x}-e^{i(m-2j)(N+1)x}}{1-e^{i(m-2j)x}}\right|\\ &\le\frac2{\left|1-e^{i(m-2j)x}\right|}\tag{3} \end{align} $$ is bounded independent of $N$, and therefore, Dirichlet's Test guarantees that $$ \sum_{n=1}^\infty\frac{e^{i(m-2j)nx}}{n^\alpha}\tag{4} $$ converges.

If $(m-2k)x\in2\pi\mathbb{Z}$, since $m$ is odd, the terms for $j=k$ and $j=m-k$ cancel in $(1)$.

Therefore, if $m$ is odd, then the original series converges.

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Case $\boldsymbol{m}$ is even

If $\sin(x)=0$, each term of the infinite sum is $0$, so the sum converges.

Suppose $\sin(x)\ne0$. If both $|\sin(nx)|$ and $|\sin((n+1)x)|$ are less than $\frac1{\sqrt2}|\sin(x)|\le\frac1{\sqrt2}$, then $|\cos((n+1)x)|\ge\frac1{\sqrt2}$, and $$ \begin{align} |\sin((n+2)x)| &=|2\sin(x)\cos((n+1)x)+\sin(nx)|\\ &\ge2|\sin(x)|\,|\cos((n+1)x)|-|\sin(nx)|\\ &\ge\left(\sqrt2-\tfrac1{\sqrt2}\right)|\sin(x)|\\ &=\tfrac1{\sqrt2}|\sin(x)| \end{align} $$ Thus, one of $|\sin(nx)|$, $|\sin((n+1)x)|$, or $|\sin((n+2)x)|$ must be at least $\frac1{\sqrt2}|\sin(x)|$. Therefore, $$ \sum_{n=1}^\infty\frac{\sin^m(nx)}{n^\alpha}\ge\left|\frac{\sin(x)}{\sqrt2}\right|^m\sum_{n=1}^\infty\frac1{(3n)^\alpha} $$ which diverges.

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Summary

The original series converges if and only if $m$ is odd or $\sin(x)=0$.

Answer 2

Let's look at the second series. To make things a little easier, suppose $0<x<1$ (The restriction that $x$ is not a rational multiple of $\pi$ is not needed.) Let $N$ be any integer such that $Nx>2\pi.$ Then among $e^{ix}, e^{i2x}, \dots , e^{iNx},$ at least one of those points lands in the arc $A=\{e^{it}: \pi/4 < t <3\pi/4\}.$ Why? Starting at $(1,0)$ we move around the unit circle in steps of arc length $1.$ One of those steps has to land in $A,$ because we traverse the entire circle at least once, and we can't step over $A$ as its arc-length is $\pi/2 > 1.$ The same holds for $(N+1)x, \dots, 2Nx$ etc.

Now the sum in question is at least

$$\tag 1 \sum_{m=1}^{\infty}\sum_{n=mN}^{(m+1)N -1}\frac{\sin^{2k}(nx)}{n}.$$

In each inner sum, we have $\sin (nx) > 2^{-1/2}$ for at least one $n.$ Thus the $m$th inner sum is at least $2^{-k}/(m+1)N.$ Hence $(1)$ is at least

$$\sum_{m=1}^{\infty}\frac{2^{-k}}{(m+1)N} = \infty.$$

So we have divergence for $x\in (0,1).$ You can tweak this argument to get divergence for $x\in (0,\pi).$ And since the series has period $\pi,$ we obtain divergence for all $x\not \in \pi \mathbb {Z}.$