It's well-known that $$\lim\limits_{n \to \infty}\frac{|\sin 1|+|\sin 2|+\cdots+|\sin n|}{n}=\frac{2}{\pi},$$which can be obtained by the uniform distribution.
Can it be used directly to solve the present problem?
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Which one? The body doesn't coincide with the title. – azif00 Jun 29 '20 at 01:16
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1@Azif00: I guess the actual question is about using $$\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\left|\sin k\right|=\frac{2}{\pi} $$ to deduce $$\lim_{n\to +\infty}\frac{1}{n^2}\sum_{k=1}^{n}k \left|\sin k\right|=\frac{1}{\pi}. $$ – Jack D'Aurizio Jun 29 '20 at 01:51
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@JackD'Aurizio Oh, sorry for the misunderstanding. – azif00 Jun 29 '20 at 01:53
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Yes, you may just invoke summation by parts. If you know that $$ s(n) = \sum_{k=1}^{n}\left|\sin k\right| = \frac{2}{\pi}n+O(1) $$ then $$ S(n) = \sum_{k=1}^{n}k\left|\sin k\right| = n s(n) - \sum_{k=1}^{n-1} s(k) $$ where $$ n s(n) = \frac{2}{\pi}n^2 + O(n), $$ $$ \sum_{k=1}^{n-1}s(k) = O(n)+\sum_{k=1}^{n-1}\frac{2}{\pi}k = \frac{1}{\pi}n^2+O(n), $$ so $$ S(n) = \frac{1}{\pi} n^2 + O(n).$$ If you start with the weaker $s(n)=\frac{2}{\pi}n+o(n)$ you end up with $S(n)=\frac{1}{\pi}n^2+o(n^2)$.
Jack D'Aurizio
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