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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

Michael Rozenberg
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math110
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  • One simple observation: $\frac{2}{\pi}$ is the average value of $|\sin k|$ over the interval $[0,2\pi]$. – Winther Sep 25 '14 at 05:26
  • Another observation: It seems that $$\lim_{n\rightarrow\infty}\left(\sum_{k=1}^n\frac{|\sin k|}{k}-\frac{2}{\pi}\log n\right)=0.610807...$$ Although I'm not sure what this number is. – karvens Sep 27 '14 at 12:10

1 Answers1

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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: $$\sum_{k=1}^n\frac{|\sin k|}{k}=\frac{2}{\pi}\log n+\frac{2}{\pi}\gamma+\mathcal{E}_n>\frac{2}{\pi}\log n+\frac{2}{\pi}\gamma>\frac{2}{\pi}\log n$$

amcalde
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