How to prove that $ \sum_{n \in \mathbb{N} } | \frac{\sin( n)}{n} | $ diverges?

Question

It is stated as a problem in Spivak's Calculus and I can't wrap my head around it.

Answer 1

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 $$\sin^2n=1-\cos^2n=1-\frac{\cos(2n)+1}2=\frac{1-\cos(2n)}2.$$

Answer 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).

Answer 3

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:

1. 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 $\frac{2}{\pi\sqrt{1-x^2}}$;

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 $\frac{2}{\pi}>0$.

Answer 4

There are many ways listed above, I will give two alternate solutions, which both rely on the following lemma :

Lemma N° 1 :: There exists an $\varepsilon > 0$ such that the set $A = \{n \in \mathbb{N} \mid |sin(n)| \geqslant \varepsilon\}$ has a strictly positive lower natural density (more precisely, that there exists $K >0$ such that $\displaystyle\frac{|A\cap[1,n]|}{n} \geqslant K$ for all $n$).

Let's show first that this lemma gives the conclusion.

Proof N°1 :

I will use the following lemma :

Lemma N°2 : if $A \subset \mathbb{N}$ is such that the series $\left(\displaystyle\frac{\mathbb{I}_A(n)}{n}\right)_{n > 0}$ is convergent then $A$ has natural density equal to $0$.

Proof of Lemma N°2 :

With the notations from the Lemma, denote by $S_n$ the quantity $S_n = \displaystyle\sum_{\sqrt{n}\leqslant k \leqslant n} \displaystyle\frac{\mathbb{I}_A(k)}{k}$. Then we have $S_n \geqslant \displaystyle\frac{A\cap[\sqrt{n},n]}{n}$. It then follows that $A$ has natural density $0$ (since there are at most $\sqrt{n}$ elements int $A\cap[1,\sqrt{n}]$).

By contrapositive, with the notations of Lemma N° 1, the series $\left(\displaystyle\frac{\mathbb{I}_A(n)}{n}\right)_{n > 0}$ is divergent. However, we have : $\displaystyle\sum_{k=1}^{n} \displaystyle\frac{|\sin(k)|}{k} \geqslant \displaystyle\sum_{k \in A_n} \displaystyle\frac{|\sin(k)|}{k} \geqslant \displaystyle\sum_{k \in A_n} \displaystyle\frac{K}{k}$ where $A_n = A\cap[1,n]$. The last term being divergent by the previous observation, we have the conclusion.

Proof N° 2 :

I will use the following results (which I will not prove for one can find it in any elementary analysis book on sequences and series).

*Result N° 1 : * If $(u_n) \in \mathbb{R_+}^{\mathbb{N}}$ is such that the series $\displaystyle\frac{u_n}{n}$ converges then $\displaystyle\lim_{n \to \infty} \displaystyle\frac{1}{n}\displaystyle\sum_{k=1}^{n} u_k = 0$ (this is know as Kronecker's lemma).

*Result N° 2 : * If $(u_n) \in \mathbb{R_+}^{\mathbb{N}}$ is such that $\displaystyle\lim_{n \to \infty} \displaystyle\frac{1}{n}\displaystyle\sum_{k=1}^{n} u_k = 0$ then there exists $B \subset \mathbb{N}$ of natural density $1$ such that $\displaystyle\lim_{n \to \infty,n \in B} u_n = 0$

The proof then follows : if the series is convergent then, by the two aforementionned results, we necessarily have that there exists $B \subset \mathbb{N}$ of natural density $1$ such that $\displaystyle\lim_{n \to \infty,n \in B} |\sin(n)| = 0$. However, with the notations of Lemma N° 1 we observe that the intersection $A\cap B$ is finite and thus the set $C = A\backslash(A\cap B)$ has a strictly positive lower natural density but is $\subset B^c$ and thus should be of natural density $0$. This is not possible.

In order to conclude, we just have to prove first lemma. I just offer here a sketch of the proof