It is stated as a problem in Spivak's Calculus and I can't wrap my head around it.
How to prove that $ \sum_{n \in \mathbb{N} } | \frac{\sin( n)}{n} | $ diverges?
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4Hint: for each $k\in\Bbb N$, the interval $[k\pi+{\pi\over 6},(k+1)\pi-{\pi\over 6}]$ contains an integer. Using this and the comparison test, show that some subseries of your series diverges. – 2012-12-25
3 Answers
Hint: $|\sin n|\geqslant \sin^2 n$, and use the convergence of $\sum_{n=1}^{+\infty}\frac{\cos(\color{red}2n)}n$ and the divergence of harmonic series.
We have $\cos(2n)=2\cos^2n-1$, and $\sin^2n=1-\cos^2n=1-\frac{\cos(2n)+1}2=\frac{1-\cos(2n)}2.$
The idea in your comment leads to one approach:
For each positive integer $k$, the interval $\bigl[k\pi+{\pi\over 6}, (k+1)\pi-{\pi\over 6}\bigr]$ has length exceeding $1$ and thus contains an integer $n_k$. We then have ${|\sin (n_k)|\over n_k} \ge {\sin({\pi\over 6})\over (k+1)\pi}$. So, from the Comparison test, it follows that $\sum\limits_{k=1}^\infty {|\sin(n_k)|\over n_k}$ diverges; whence $\sum\limits_{n=1}^\infty {|\sin(n )|\over n}$ diverges (by the Comparison test again).
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1@David Mitra: how did you choose this specific interval($[\pi k +\frac{\pi}{6},\pi(k+1)-\frac{\pi}{6}$? – 2013-06-23
A slightly different approach. Assuming that $\sum_{n\geq 1}\frac{\left|\sin n\right|}{n}$ is convergent, Kronecker's lemma ensures that $\{\left|\sin n\right|\}_{n\geq 1}$ has mean zero. Now there are at least a couple of ways (besides $\left|\sin n\right|\geq \sin^2 n$) for showing this is absurd:
By the equidistribution theorem, $e^{in}$ essentially is a random point on $S^1$ for $n\in\mathbb{N}$, hence $\left|\sin n\right|$ essentially behaves like a random variable, whose density is supported on $(0,1)$ and given by $\frac{2}{\pi\sqrt{1-x^2}}$;
The Fourier cosine series of $\left|\sin x\right|$ is given by $ \left|\sin x\right|=\color{red}{\frac{2}{\pi}}-\frac{4}{\pi}\sum_{m\geq 1}\frac{\cos(2mx)}{4m^2-1} $ hence the average value of $\{\left|\sin n\right|\}_{n\geq 1}$ is $\frac{2}{\pi}>0$.