Show the divergence of the series $\sum \frac{\sin^2n}{n}$ without Dirichlet's test

Question

Show that the series $$ \sum_{n\in\mathbb N} \frac{\sin^2n}{n} $$ is divergent.

I know how to do this with the Dirichlet's test. But is there any other way to prove it? Thanks!

You could observe $\sum \frac{\sin^2(n)}{n} = \sum \frac{1}{2n} - \sum \frac{\cos(2n)}{2n}$ and then apply to $\sum \frac{\cos(2n)}{2n}$ a similar argument as https://math.stackexchange.com/questions/13490/proving-that-the-sequence-f-nx-sum-limits-k-1n-frac-sinkxk-is? — user, Jan 08 '18 at 10:38
@gimusi It involves Fourier series? However, I'm newbie on that... But, anyway, that's an interesting post and I will read carefully later. — Macrophage, Jan 08 '18 at 10:40
@gimusi Actually I know how to reduce power of sin, just look for another way to prove convergence of sin(nx) for x!=2kpi — Macrophage, Jan 08 '18 at 10:42
I haven't found others references here whitout Dirichlet's test. Here the Abel partial summation has been used but it seems a generalization of Dirichlet https://math.stackexchange.com/questions/270057/give-a-demonstration-that-sum-limits-n-1-infty-frac-sinnn-converges?noredirect=1&lq=1 — user, Jan 08 '18 at 10:47
@gimusi Well, then I shall take some time reading the post you provided.... Thanks :D — Macrophage, Jan 08 '18 at 10:48
@gimusi It's the contrary, a specialization, while using Abel partial summation to prove Dirichlet's test is the more general case. Of course, one could do that, but it's a waste of time and space without any new insight. — , Jan 08 '18 at 10:56
@ProfessorVector Thanks, indeed I only gave a quick sight to the post. What is the best way without Dirichlet to prove the convergence of the remainder term $\sum \frac{\cos(2n)}{2n}$? — user, Jan 08 '18 at 10:58
@gimusi I wouldn't use anything except Abel partial summation, it does the job just nicely. You could use general results about the remainder of a Fourier series, but that wouldn't be simpler, and guess which mathematician obtained those results... — , Jan 08 '18 at 11:05
The mean value of $\sin^2(n)$ is $\frac{1}{2}$, hence the given series is divergent by Kronecker's Lemma. — Jack D'Aurizio, Jan 08 '18 at 11:27
@ProfessorVector I've tried to give an answer by your suggestion, can it work? Thanks — user, Jan 08 '18 at 11:28
@JackD'Aurizio Hence there is a shorcut! I'll take a look to it, thanks! — user, Jan 08 '18 at 11:30
It's not a coincidence that the proof of Kronecker's Lemma uses Abel partial summation. — , Jan 08 '18 at 12:19
Of any three consecutive values of $\sin^2 n$, at least one is $\geqslant \frac{1}{4}$. Hence $$\sum_{n = 1}^{3k} \frac{\sin^2 n}{n} \geqslant \sum_{m = 1}^k \frac{1}{4\cdot 3k} > \frac{1}{12}\log k,.$$ — Daniel Fischer, Jan 08 '18 at 13:32
@DanielFischer This morning I've read about this result in some other answer to related OP, but I can't find out it now. I can't see it is true immediately, how can we prove it? Thanks. — user, Jan 08 '18 at 14:10
@gimusi If $\lvert \sin n\rvert < \frac{1}{2}$, then there is a $k$ with $\bigl(k - \frac{1}{6}\bigr)\pi < n < \bigl(k + \frac{1}{6}\bigr)\pi$. Then $\bigl(k+\frac{1}{6}\bigr)\pi < \bigl(k - \frac{1}{6}\bigr)\pi + 2 < n+2 < \bigl(k+\frac{1}{6}\bigr)\pi + 2 < \bigl(k + \frac{5}{6}\bigr)\pi$, whence $\lvert \sin (n+2)\rvert \geqslant \frac{1}{2}$. — Daniel Fischer, Jan 08 '18 at 14:20
On a general note: The answer to the question "But is there any other way to prove it?" Is always "yes", regardless of what "it" is (provided there is indeed some way to prove it). This is particularly true when the given way of proving is by some special test, as here. After all, these tests were themselves proved from the definitions. — Paul Sinclair, Jan 08 '18 at 14:53
@Macrophage Please, if you are ok, you can accept the answer and set it as solved. Thanks! — user, Feb 03 '18 at 23:59

user · Accepted Answer · 2018-01-08T13:49:22.333

4

Note that

$$\sum_{n=1}^{\infty} \frac{\sin^2(n)}{n} = \sum_{n=1}^{\infty} \frac{1}{2n} - \sum_{n=1}^{\infty} \frac{\cos(2n)}{2n}$$

which diverges since

$\sum \frac{1}{2n}$ diverges

$\sum \frac{\cos(2n)}{2n}$ converges

indeed by Abel transformation and Lagrange's trigonometric identities, let

$$a_n=\frac{1}{2n} \quad b_n=\cos(2n)=B_{n}-B_{n-1}\quad B_n=\sum_{k=1}^{n} \cos (2k)= -\frac12+\frac{\sin(2n+1)}{2\sin 1}$$

$$S_N=\sum_{n=1}^{N}a_nb_n=\sum_{n=1}^{N} \frac{\cos(2n)}{2n} =\frac{1}{2N}\left(-\frac12+\frac{\sin(2N+1)}{2\sin 1}\right)-\sum_{n=1}^{N-1} \left[ \left(-\frac12+\frac{\sin(2n+1)}{2\sin 1}\right)\left(\frac{1}{2n+2}-\frac{1}{2n}\right)\right]=$$

$$=\frac{1}{2N}\left(-\frac12+\frac{\sin(2N+1)}{2\sin 1}\right)-\sum_{n=1}^{N-1} \frac{1}{4n(n+1)}+\sum_{n=1}^{N-1} \frac{\sin (2n+1)}{4n(n+1)\sin 1}$$

and

$$\sum_{n=1}^{\infty} \frac{\sin (2n+1)}{4n(n+1)\sin 1}$$

converges absolutely by comparison with $\sum \frac{1}{n^2}$.

edited Jan 08 '18 at 13:49

answered Jan 08 '18 at 11:27

user

154,566

It is still wrong: the Abel transformation means that you have differences multiplied by partial sums of $\cos2n$. Those are still bounded, that's why Dirichlet's test works, but you have to show that. – Jan 08 '18 at 12:23
@ProfessorVector Seems tricky... I shall wait for some more edits to this answer. – Macrophage Jan 08 '18 at 12:26
@ProfessorVector Thanks, I'll take a look accordingly to your suggestion! – user Jan 08 '18 at 12:39
@ProfessorVector I've revised the solution, does it work now? Thanks. – user Jan 08 '18 at 13:10
1

Yes, it's ok, now. – Jan 08 '18 at 13:13
@ProfessorVector Thanks a lot for the guide and patience. It's a pleasure for me can learn all this things by persons skilled and expert like you. – user Jan 08 '18 at 13:18
1

Minor quibble: Yours sums should start at $n=1$, not $n=0$. – Barry Cipra Jan 08 '18 at 13:44
@BarryCipra Yes of course! Thanks – user Jan 08 '18 at 13:45

achille hui · Answer 2 · 2018-01-08T14:22:03.850

For any $a, b \in \mathbb{R}^n$, we have

$$\begin{align}\sin^2(a) + \sin^2(b) &= \frac{1-\cos(2a)}{2} + \frac{1-\cos(2b)}{2} &= 1 - \frac12(\cos(2a)+\cos(2b))\\ = 1 - \cos(a+b)\cos(a-b) \end{align} $$ For any integer $k$, this leads to

$$\sin^2(2k-1) + \sin^2(2k) = 1 - \cos(4k-1)\cos(1) \ge 1-\cos(1) > 0$$

As a result, we can bound the partial sums of even number of terms from below as

$$\sum_{n=1}^{2p}\frac{\sin^2(n)}{n} = \sum_{k=1}^{p}\left(\frac{\sin^2(2k-1)}{2k-1} + \frac{\sin^2(2k)}{2k}\right) \ge \sum_{k=1}^p \frac{1 - \cos(1)}{2k} = \frac{1-\cos(1)}{2}H_p$$

Since the harmonic number $H_p$ diverges like $\log(p)$ for large $p$, the partial sums of even number of terms diverges to $\infty$. As a consequence, the sum $\displaystyle\;\sum_{n=1}^\infty \frac{\sin^2(n)}{n}$ diverges.

score 2 · Answer 3 · answered Jan 08 '18 at 13:54

2

If you march around the unit circle in strides of radian $1$, you find that $|\sin n|\gt\sqrt3/2$ at least once for every six steps. Thus

$$\sum{\sin^2n\over n}\gt\sum{3\over4(6k)}=\sum{1\over8k}=\infty$$

(Ah, I see that Daniel Fischer gave essentially the same answer as a comment.)

answered Jan 08 '18 at 13:54

Barry Cipra

79,832

This morning I've read about this result in some other answer to related OP, but I can't find out it now. I can't see it is true immediately, how can we prove it? Thanks. – user Jan 08 '18 at 14:07

Show the divergence of the series $\sum \frac{\sin^2n}{n}$ without Dirichlet's test

3 Answers3