Is there a tighter upper bound for $\sum_{k=1}^n|k\sin k|$ than $\frac12n(n+1)$? Consider the sum
$$\sum_{k=1}^n|k\sin k|$$
An obvious upper bound for this is clearly $\sum_{k=1}^n k=\frac12n(n+1)$. But it seems that this upper bound is too "loose", and so I was wondering if it is possible to find a tighter upper bound for it?
Below please find a Mathematica plot for the sum in the interval $n\le1000$.

 A: The average value of $\left|\sin k\right|$ is $2/\pi$ so an upper bound is probably near $$\frac{n(n+1)}{\pi}$$
As the plot below shows, this is quite accurate when $900\le n\le1000$. $n(n+1)/\pi$ is the red curve, the scatterplot im blue represents the actual sums.

At most one in every three of $|\sin k|$ is more than $\cos 0.5$ so one bound is near $$\frac{1+2\cos0.5 }6n(n+1)$$
In the same way, the average of $22$ consecutive values of $|\sin k|$ is always between $0.635$ and $0.638$.  So the sum is bounded above by
$$\sum_{k=1}^N k\left|\sin k\right|\lt 0.638\sum_{k=1}^N \left(N-22\lfloor (N-k)/22 \rfloor\right)$$
where each $k$ has been rounded up to the nearest number of the form $N-22m$.
This has a polynomial sum if $N$ is a multiple of $22$, and a finite correction if not.
$$\sum_{k=1}^N k\left|\sin k\right|\lt 0.638\frac{N(N+22)}2+C$$
By noting that $\sum_{k=1}^{22}k|\sin(M+k)|$ is always between $153$ and $169$, this can be improved to
$$0.3175N(N-0.1)+C_1\lt\sum_{k=1}^N k\left|\sin k\right|\lt0.319N(N+2)+C_2$$
