Value of $\sum_{n=1}^\infty\sin^{2k+1}(n)/n$

Question

Prove or disprove: for any positive integer $k$, $$\sum_{n=1}^\infty\frac{\sin^{2k+1}(n)}{n}=\frac{1}{3}\left(\frac{3}{4}\right)^k\pi$$ This is a conjecture, which I haven't found anywhere else. I've checked it using Wolfram Alpha for small values of $k$ (less than 7), but I'm not sure how to prove it in general. I think complex analysis is needed, but I'm a bit rusty on this.

Answer 1

Disproof: It suffices to set $k=12$, then \begin{align} \tilde{S}_{12}=\frac13\left(\frac34\right)^{12}\pi=&\frac{\color{red}{177\,147}\,\pi}{16\,777\,216},\\ S_{12}=\sum_{n=1}^{\infty}\frac{\sin^{25}n}{n}=&\frac{\color{blue}{177\,146}\,\pi}{16\,777\,216}. \end{align} Addendum: The exact formula for the sum is $$S_k=\sum_{n=1}^{\infty}\frac{\sin^{2k+1}n}{n}=2^{-2k}\sum_{j=0}^k\left(-1\right)^{k-j}{2k+1\choose j}\left[\frac{\pi}{2}-\pi\left\{\frac{2k-2j+1}{2\pi}\right\}\right],\tag{1}$$ where $\{x\}$ denotes the fractional part of $x$. This formula is a consequence of two basic observations:

1. Using binomial expansion, one can express $\sin^{2k+1}n$ as a linear combination of sines of multiple angles: \begin{align} \sin^{2k+1}n&=\left(\frac{e^{in}-e^{-in}}{2i}\right)^{2k+1}=2^{-2k}(-1)^k\sum_{j=0}^{2k+1}{2k+1 \choose j}(-1)^j \frac{e^{i(2k+1-2j)n}}{2i}=\\&= 2^{-2k}\sum_{j=0}^{k}{2k+1 \choose j}(-1)^{k-j}\sin\left(2k-2j+1\right)n.\tag{2} \end{align}

2. Consider the periodic function $f(x)$ of period $2\pi$ defined by $$f(x)=\frac{\pi-x}{2},\qquad x\in[0,2\pi).$$ A convenient way to represent this function is to write $f(x)=\frac{\pi}{2}-\pi\left\{\frac{x}{2\pi}\right\}$ for $x\ge 0$. On the other hand, this function can be easily expanded into Fourier series, which yields the identity $$\frac{\pi}{2}-\pi\left\{\frac{x}{2\pi}\right\}=\sum_{n=1}^{\infty}\frac{\sin nx}{n},\qquad x\in\mathbb{R}_{>0}\backslash 2\pi\mathbb{Z}_{>0}.\tag{3}$$ The integer multiples of $2\pi$ are excluded since at these points $f(x)$ is discontinuous and does not have to coincide with its Fourier series. This is however of no importance for our problem where the relevant $x$'s are integer.

Combining (2) and (3) immediately gives (1).

Starting from $k=12$ the sum (1) differs from the conjectured expression. Moreover the difference seems to increase with the growth of $k$ and tend to some finite value as $k\to \infty$.

Answer 2

I suppose that this is coming from the fact that$$\sin^{2m+1}(x)=\frac 1{2^{2m}}\sum_{k=0}^{m}(-1)^{m-k}\binom{2m+1}{k}\sin\Big((2m-2k+1)x\Big)$$ and that $$S_{k}=\sum_{n=1}^\infty \frac{\sin\big((2k+1) n\big)}n=\alpha_k\,\pi+\beta_k $$ where $$\alpha_k=\lfloor \frac{k-1}{3}\rfloor +\frac{1}{2}$$ $$\beta_k=-\big(k+\frac{1}{2}\big)$$

I hope and wish this will help you for the beautiful conjecture.

Answer 3

Let $i=\sqrt{-1}$. \begin{align} f(k)&=\sum_{n=1}^\infty\frac{\sin^{2k+1}(n)}{n}\\ &=\sum_{n=1}^\infty\frac1n\frac{i^{2k+1}}{2^{2k+1}}\sum_{j=0}^{2k+1}\binom{2k+1}{j}(-1)^{2k+1-j}e^{in(2j-2k-1)}\\ =&\frac{i^{2k+1}}{2^{2k+1}}\sum_{j=0}^{2k+1}(-1)^{2k+1-j}\binom{2k+1}{j}\sum_{n=1}^\infty \frac1n e^{in(2k+1-2j)}\\ =&\frac{i^{2k+1}}{2^{2k+1}}\sum_{j=0}^{2k+1}\binom{2k+1}{j}(-1)^{2k+1-j}\Big(-\log(1-e^{i(2k+1-2j)})\Big) \end{align}

I did not manage to simplify this to your neat closed form result!