Let $p>1$ , $p\in\mathbb{R}$. For what values of $p$, the series $\sum_{n=1}^{\infty}|\frac{\sin(n)}{n}|^p$ is convergent?
When $p=1$, I know the series is divergent but how about other cases? Thanks!
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If $p > 1$ we can write:
$0 < \sum\limits_{n=1}^{\infty}|\frac{\sin(n)}{n}|^p \le \sum\limits_{n=1}^{\infty}|\frac{1}{n}|^p$
and the right part converges.So the middle part also converges as $\sum\limits_{n=1}^{k}|\frac{\sin(n)}{n}|^p$ is an increasing function of $k$ and it's limited by $\sum\limits_{n=1}^{\infty}|\frac{1}{n}|^p$.
So for every $p > 1$ it converges.
Marco Cantarini
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Andrei Kulunchakov
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That should be $p>1$, not $p \geq 1$ – ASKASK Oct 28 '15 at 06:45
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How do we know $\sum\limits_{n=1}^{\infty}\frac{1}{n^p}$ is convergent when p>1$ – Ergin Süer Oct 28 '15 at 06:47
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@ErginSuer: See http://math.stackexchange.com/questions/29450/self-contained-proof-that-sum-limits-n-1-infty-frac1np-converges-for – Oct 28 '15 at 06:49
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I suppose that there you can find some proofs http://math.stackexchange.com/questions/367135/why-does-the-harmonic-series-diverge-but-the-p-harmonic-series-converge – Andrei Kulunchakov Oct 28 '15 at 06:49