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

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 math.stackexchange.com/questions/13490/…? – gimusi Jan 8 '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 8 '18 at 10:40
• 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 math.stackexchange.com/questions/270057/… – gimusi Jan 8 '18 at 10:47
• 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 8 '18 at 13:32
• @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 8 '18 at 14:20

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 $\sum \frac{1}{n^2}$.

• 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. – Professor Vector Jan 8 '18 at 12:23
• @ProfessorVector Seems tricky... I shall wait for some more edits to this answer. – Macrophage Jan 8 '18 at 12:26
• @ProfessorVector Thanks, I'll take a look accordingly to your suggestion! – gimusi Jan 8 '18 at 12:39
• Yes, it's ok, now. – Professor Vector Jan 8 '18 at 13:13
• Minor quibble: Yours sums should start at $n=1$, not $n=0$. – Barry Cipra Jan 8 '18 at 13:44

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.

• Very nice proof! – gimusi Jan 8 '18 at 14:09

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.)

• 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. – gimusi Jan 8 '18 at 14:07