TITLE: Evaluate $\lim\limits_{n \to \infty}\sum_{k=n+1}^{2n}\frac{|\sin 1|+2|\sin 2|+\cdots+k|\sin k|}{k^3}$.
QUESTION [6 upvotes]: Evaluate

$$\lim\limits_{n \to \infty}\sum_{k=n+1}^{2n}\frac{|\sin  1|+2|\sin
 2|+\cdots+k|\sin k|}{k^3}.$$

At least, we can estimate an upper bound as follows:
\begin{align*}
\sum_{k=n+1}^{2n}\frac{|\sin
1|+2|\sin 2|+\cdots+k|\sin k|}{k^3}&\le\sum_{k=n+1}^{2n}\frac{1+2+\cdots+k}{k^3}\\&=\sum_{k=n+1}^{2n}\frac{k+1}{2k^2}\\&=\frac{1}{2}\sum_{k=n+1}^{2n}\frac{1}{k}+\frac{1}{2}\sum_{k=n+1}^{2n}\frac{1}{k^2}\\&\to \frac{1}{2}\ln 2(n \to \infty).
\end{align*}
But how to obtain the lower bound?

REPLY [6 votes]: $$\left|\sin x\right| = \frac{2}{\pi}-\sum_{m\geq 1}\frac{4}{\pi(4m^2-1)}\cos(2m x) =\frac{2}{\pi}-f(x)$$
where $f(x)$ is a Lipschitz-continuous, $\pi$-periodic, even function with mean zero. It follows that
$$ \sum_{k=n+1}^{2n}\frac{1}{k^3}\sum_{j=1}^{k}j\left|\sin j\right|=\frac{2}{\pi}\sum_{k=n+1}^{2n}\frac{k+1}{2k^2}-\sum_{k=n+1}^{2n}\frac{1}{k^3}\sum_{j=1}^{k}j\cdot f(j) $$
and by letting $F(m)=f(1)+f(2)+\ldots+f(m)$ and invoking summation by parts we have
$$ \sum_{k=n+1}^{2n}\frac{1}{k^3}\sum_{j=1}^{k}j\left|\sin j\right|=\frac{\log 2}{\pi}+o(1)-\sum_{k=n+1}^{2n}\frac{1}{k^3}\left(k F(k)-\sum_{j=1}^{k-1}F(j)\right) $$
and the claim
$$ \lim_{n\to +\infty}\sum_{k=n+1}^{2n}\frac{1}{k^3}\sum_{j=1}^{k}j\left|\sin j\right|=\frac{\log 2}{\pi} $$
simply follows from the boundedness of $F(x)$, granted by Weyl's criterion.
Actually we just need $F(x)=o(x)$, which is fairly obvious given the premises.