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I know that the series $\sum_{k=1}^\infty \frac{\sin k}{k}$ converges (to $\frac{\pi - 1}{2}$), though by crazy stuff with Dirichlet Kernels or by reverse-engineering $\frac{\pi - x}{2} = \sum \frac{\sin (kx)}{k}$ using Fourier Series.

My question instead: Is $\sum_{k=1}^\infty \frac{\sin k}{k}$ absolutely convergent?

My gut instinct is no: It behaves not-much-better than $\sum \frac1k$, but that's certainly not a rigorous proof.

Martin Sleziak
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D.J.
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1 Answers1

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We need to investigate whether $\sum_{k=1}^\infty \frac{|\sin k|}{k}$ convergent.

Note $\forall k\in\mathbb{N},[\frac{\pi}{6}+2k\pi,\frac{5\pi}{6}+2k\pi]$,$\frac{5\pi}{6}+2k\pi-(\frac{\pi}{6}+2k\pi)=\frac{4\pi}{6}>1$, hence there exists $n_k\in\mathbb{N},n_k\in[\frac{\pi}{6}+2k\pi,\frac{5\pi}{6}+2k\pi]$. Hence $|\sin n_k|>\frac{1}{2}$, and $\frac{1}{n_k}\geq \frac{1}{\frac{5\pi}{6}+2k\pi}>\frac{1}{(1+2k)\pi}$.

$\forall k\in\mathbb{N}$, we find such $n_k$, then $$\sum_{k=1}^\infty \frac{|\sin k|}{k}>\sum_{k=1}^\infty \frac{|\sin n_k|}{n_k}>\sum_{k=1}^\infty\frac{1}{(1+2k)2\pi}$$

the RHS series is divergent, by comparison test with hamonic series. So the series is not absolutely convergent.

John
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  • Aha, that's very sleek and definitely rigorous! Thanks!

    (Minor note on the last line: Should be RHS divergent by comparison.)

     – D.J. Nov 04 '14 at 15:49
  • @D.J. You are right. Sorry for the typo. I will edit my answer. – John Nov 04 '14 at 16:02
  • @john Can u please tell how to prove this.$\sum_{k=1}^\infty \frac{|\sin k|}{k}>\sum_{k=1}^\infty\frac {|sinn_k|}{n_k} $ – sajan Oct 17 '18 at 03:36