Question about $\sum_{n=1}^{\infty}\frac{|\sin(n)|}{n}$.

Question

In several places on this site the sum $\displaystyle\sum_{n=1}^{\infty}\dfrac{\sin(n)}{n}$ has been discussed as a generalized alternating series, which therefore converges. I am curious about the sum $\displaystyle\sum_{n=1}^{\infty}\dfrac{|\sin(n)|}{n}$, which I am sure does not converge. Here is my "proof". Steps 2 and 3 clearly needs some clarification.

Step 1) Define $A\subset\mathbb{R}$ by $A:=[\pi/4,3\pi/4]\cup[5\pi/4,7\pi/4]\cup\ldots$. For any $x\in A$, $|\sin(x)|\geq\sqrt{2}/2$.

Step 2) Since $\pi$ is irrational, roughly half the integers are in $A$. That is, if we define $\#_A(n)$ to be the cardinality of $\{1,\ldots,n\}\cap A$, $\displaystyle\lim_{n\to\infty}\dfrac{\#_A(n)}{n}=\dfrac{1}{2}$.

Step 3) Therefore $\displaystyle\sum_{n=1}^{\infty}\dfrac{|\sin(n)|}{n}>\sum_{n>0,n\in A}\dfrac{|\sin(n)|}{n}>\sum_{n>0,n\in A}\dfrac{\sqrt{2}/2}{n}\approx\dfrac{1}{2}\sum_{i=1}^n\dfrac{\sqrt{2}/2}{n}$.

Step 4) Since the last sum in Step 3 diverges, the sum $\displaystyle\sum_{n=1}^{\infty}\dfrac{|\sin(n)|}{n}$ diverges.

I beleive this is more or less intuitively right, but I do not really know the theory to make rigorous Steps 2 and 3.

Answer 1

Your idea is almost correct, except that we need some justification. So here is the missing part:

Lemma. If $(a_{n})_{n=1}^{\infty}$ be such that $\sigma_{n} = (a_{1}+\cdots+a_{n})/n$ converges to $\alpha > 0$, then $$ \sum_{n=1}^{\infty} \frac{a_{n}}{n} = \infty. \tag{1}$$

Proof. Let $s_{n} = a_{1} + \cdots + a_{n}$ denotes the partial sum. Then for $n \geq 2$,

$$ \frac{a_{n}}{n} = \sigma_{n} - \sigma_{n-1} + \frac{\sigma_{n-1}}{n}. $$

Consequently, we have

$$ \sum_{n=1}^{N} \frac{a_{n}}{n} = \sigma_{N} + \sum_{n=1}^{N-1} \frac{\sigma_{n}}{n+1}. $$

Using standard techniques, it is easy to see that

$$ \frac{1}{\log N} \sum_{n=1}^{N-1} \frac{\sigma_{n}}{n+1} \xrightarrow[]{N\to\infty} \alpha. $$

Thus we have

$$ \sum_{n=1}^{N} \frac{a_{n}}{n} = (\alpha+o(1))\log N $$

and hence (1) follows. ////

Answer 2

$f(x)=x;- \pi \le x \le \pi$

$x \in \left[ { - \pi ;\pi } \right]; - x \in \left[ { - \pi ;\pi } \right]:f( - x) = - f(x)$

$\forall n \in N:a_n= 0$

$b_n = \frac{2}{\pi }\int\limits_0^\pi {x\sin nxdx = } - \frac{2}{{\pi n}}\left[ {x\cos nx} \right]_0^\pi$

$ b_n = - \frac{2}{{n\pi }}\left[ {x\cos nx} \right]_0^\pi = \frac{2}{n}\left( { - 1} \right)^{n + 1}$

$f(x) = 2\sum\limits_{n = 1}^{ + \infty } {\frac{{\left( { - 1} \right)^{n + 1} \sin nx}}{n}} $

$x\in\left] { - \pi ;\pi } \right]:\frac{x}{2} = \sum\limits_{n = 1}^{ + \infty } {\left( { - 1} \right)^{n + 1} \frac{{\sin nx}}{n}}$

$ t \in \left] {0;2\pi } \right]:\left( {\pi - t} \right) \in \left] { - \pi ;\pi } \right]$

if pose $x = \pi - t$

$\displaylines{ \sin \left( {n\left( {\pi - t} \right)} \right) = \cos n\pi \sin \left( { - nt} \right) + \sin n\pi \cos \left( { - nt} \right) \cr = \left( { - 1} \right)^n \sin \left( { - nt} \right) = \left( { - 1} \right)^{n + 1} \sin \left( {nt} \right) \cr}$

$\sum\limits_{n = 1}^{ + \infty } {\frac{{\sin nt}}{n}} = \frac{{\pi - t}}{2};\forall t \in \left] {0;2\pi } \right]$ t=1 $\sum\limits_{n = 1}^{ + \infty } {\frac{{\sin n}}{n}} = \frac{{\pi - 1}}{2}$

Answer 3

You could use the fact that the intervals $\left(\frac{(4n-1)\pi}{4},\frac{(4n+1)\pi}{4}\right)$ have length $\frac{\pi}{2}<2$,

so there can't be 3 consecutive integers in any of these intervals;

so $S_{3n}=\frac{|\sin 1|}{1}+\frac{|\sin 2|}{2}+\frac{|\sin3|}{3}+\cdots+\frac{|\sin 3n|}{3n}\ge\frac{\sqrt{2}}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\cdots+\frac{1}{3n}\right)\to\infty$.

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Another argument you could use is that

$\displaystyle\frac{|\sin n|}{n}\ge\frac{\sin^2 n}{n}=\frac{\frac{1}{2}(1-\cos 2n)}{n}\ge0$ for all n, and

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ diverges while $\displaystyle\sum_{n=1}^{\infty}\frac{\cos2n}{n}$ converges by Dirichlet's test,

so $\displaystyle\sum_{n=1}^{\infty}\frac{|\sin n|}{n}$ diverges by the Comparison Test.