Does $\sum \frac{\sin^k j}{j}$ converge or diverge where $k$ is a positive integer? I know that for $k=1$ the series converges and for $k=2$ the series diverges. What can we say about its convergence when $k>2$? The usual proof for $k=2$ is to reduce the summand to $\frac{1-\cos\left( 2j \right)}{2j}$ using a trig identity, but it's not clear to me how this can be generalized to $k>1$.
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(https://math.stackexchange.com/questions/273559/convergence-of-sum-n-1-infty-frac-sin2nn) you can apply similar trick from the answer to prove it diverges for even $k$, intuitively I'd assume it converges for odd $k$ but I have no idea how to prove it (if it's correct). – kingW3 Oct 17 '19 at 03:24
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what if you write $\sin^{2k+1}j/j = \sin^{2k-1}j/j - \cos^2j\sin^{2k-1}j/j ?$ – dezdichado Oct 17 '19 at 03:56
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It converges for odd values of $k$ and it diverges for even values of $k$.
Indeed, if $k$ is even then $\sin^k(n)$ has a positive mean value and the divergence is a consequence of Kronecker's lemma.
If $k$ is odd then $\sin^k(n)$ can be written as a linear combination of $\sin(n),\sin(3n),\sin(5n)$ etcetera, and $$ \sum_{n\geq 1}\frac{\sin(nx)}{n} $$ equals the $2\pi$-periodic extension of the function which equals $\frac{\pi-x}{2}$ on $(0,2\pi)$.
Examples:
$$\sum_{n\geq 1}\frac{\sin^3(n)}{n}=\frac{3}{4}\sum_{n\geq 1}\frac{\sin(n)}{n}-\frac{1}{4}\sum_{n\geq 1}\frac{\sin(3n)}{n}=\frac{3}{4}\cdot\frac{\pi-1}{2}-\frac{1}{4}\cdot\frac{\pi-3}{2}=\color{red}{\frac{\pi}{4}}.$$ Since $\sin^4(x)=\frac{3}{8}-\frac{1}{2}\cos(2x)+\frac{1}{8}\cos(4x)$, $$ \sum_{n=1}^{N}\frac{\sin^4(n)}{n} \geq \frac{3}{8}H_n - C.$$
Jack D'Aurizio
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How did you get the expansion for $\sin^3$? Is there a formula you used? – Aden Dong Oct 17 '19 at 15:00
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@AdenDong: you may just play with $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and the binomial theorem. – Jack D'Aurizio Oct 17 '19 at 15:38
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