How prove this inequality $\sum\limits_{k=1}^{n}\frac{|\sin{k}|}{k}>\frac{2}{\pi}\ln{n}$ Question:

show that
  $$\sum_{k=1}^{n}\dfrac{|\sin{k}|}{k}>\dfrac{2}{\pi}\ln{n}$$

I know this 
$$\sum_{k=1}^{\infty}\dfrac{\sin{k}}{k}=\dfrac{\pi-1}{2}$$
But for this inequality,I can't.Thank you
 A: For large $n$ this is true because as @Winther said, $|\sin k|$ averages to $2/\pi$, and since the integers are incommensurate with $\pi$, they'll be ergodic in $[0,2\pi)$ so
$$[1]\qquad\qquad\qquad \lim_{m \rightarrow \infty}\sum_{k=m}^n\frac{|\sin k|}{k}=\frac{2}{\pi}\sum_{k=m}^n\frac{1}{k}=\frac{2}{\pi}(H_n-H_{m-1})$$where $H_n$ is the $n$th harmonic sum. 
Also $H_n = \log n + \gamma +\mathcal{O}(\frac{1}{n})$ so we have 
$$[2]\qquad \sum_{k=1}^n\frac{|\sin k|}{k}=\frac{2}{\pi}H_n+\left(\sum_{k=1}^n\frac{|\sin k|-(2/\pi)}{k}\right)= \frac{2}{\pi}\log n+\frac{2}{\pi}\gamma+\mathcal{E}_n$$ where $\gamma$ is the Euler–Mascheroni constant, and the error terms (from averaging $|\sin|$ and summing the harmonic sequence) are grouped into $\mathcal{E}_n$. From $[1]$ above, we have shown that the differences in successive error terms go to zero, so the only thing left to check is if they are always greater than $-2 \gamma/\pi$ for every $n$. In fact for every $n$ we have $$ 0 < \mathcal{E}_n $$ Since the error terms are always positive, we easily have:
$$\sum_{k=1}^n\frac{|\sin k|}{k}=\frac{2}{\pi}\log n+\frac{2}{\pi}\gamma+\mathcal{E}_n>\frac{2}{\pi}\log n+\frac{2}{\pi}\gamma>\frac{2}{\pi}\log n$$
