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I want to detemine the convergence of the next series:

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

I've solved the limit: $$\lim_{n->\infty}\frac{\sin(na)}{n^2}=\frac{[ -1,1]}{\infty}=0$$ The series has the necesary condiction of convergence (limit=0), but I dont go further from here

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4 answers all of similar content within a 20 second posting span. That's cool =]. –  Ragib Zaman Jun 17 '12 at 14:58
@RagibZaman Yeah, it is. The only problem is that I can't shake the feeling that the OP meant $\sum\frac{\sin(n\alpha)}{n}$. –  dtldarek Jun 17 '12 at 15:08
Dividing an interval by infinity. I never seen that! (JK: Careful with the notation, you should write that more rigorously, use the sandwich theorem, for example.) –  Pedro Tamaroff Jun 17 '12 at 15:12
@dtldarek: I have the same feeling! Fortunately, it's simply an aesthetic difference. –  Cameron Buie Jun 17 '12 at 15:41
This is a duplicate. Its abstract duplicate $\sum \dfrac{\sin(nx)}{n^k}$ was discussed here few weeks back. –  user17762 Jun 17 '12 at 16:24
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5 Answers

up vote 3 down vote accepted

Use the comparison test. Since $\left|\sin(na)\right| \le 1$, we have:

$$ \left| \frac{\sin(na)}{n^2} \right| \le \left| \frac{1}{n^2} \right| $$

We know that $\sum_{n=1}^\infty \frac{1}{n^2}$ converges absolutely, so your series does too.

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Show that it’s absolutely convergent, i.e., that $$\sum_{n=1}^\infty\left|\frac{\sin na}{n^2}\right|$$ converges. You should have no trouble doing this by comparing it with a simple series that you know converges.

Note, by the way, that it makes no sense at all to write $$\lim_{n\to\infty}\frac{\sin na}{n^2}=\frac{[-1,1]}\infty\;:$$ $\infty$ isn’t something by which you can divide, and $[-1,1]$ isn’t even a number. What you mean is this:

For all $n$, $$\frac{-1}{n^2}\le\frac{\sin na}{n^2}\le\frac1{n^2}\;,$$ so $$0=\lim_{n\to\infty}\frac{-1}{n^2}\le\lim_{n\to\infty}\frac{\sin na}{n^2}\le\lim_{n\to\infty}\frac1{n^2}=0\;,$$ and therefore by the squeeze theorem $$\lim_{n\to\infty}\frac{\sin na}{n^2}=0\;.$$

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Commented before seeing this. We concur, Mr. Scott. –  Pedro Tamaroff Jun 17 '12 at 15:13
@PeterTamaroff Probably best to call him by his proper title, either Dr. or Professor. I've seen Dr. Pete Clark feel disrespected by this, so that probably means others would as well. –  Ragib Zaman Jun 17 '12 at 15:30
@Ragib: I’m not fussy: my first name is fine. As far as titles go, I agree with my father: when asked whether he should be called Doctor or Professor, he said, ‘ Mister is always correct’. –  Brian M. Scott Jun 17 '12 at 15:33
@RagibZaman I thought Brian was his first name, Scott his last. –  Pedro Tamaroff Jun 17 '12 at 15:36
@PeterTamaroff Indeed. I was referring to the "Mr." part, but as Brian has told us, it's all the same to him =). –  Ragib Zaman Jun 17 '12 at 15:37
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$-1 \leq \sin(na) \leq 1$ and hence

$$\left| \frac{sin(na)}{n^2} \right| \leq \frac{1}{n^2}$$

Since $\sum \frac{1}{n^2}$ is convergent....

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We get absolute convergence by the comparison test since $$ \sum_{n=1}^{\infty} |\frac{\sin na}{n^2}| \leq \sum_{n=1}^{\infty} \frac{1}{n^2}, $$ and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges. Since the series converges absolutely, it converges. This holds for every real number $a$.

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We can use direct comparison of $$\sum_{n=1}^\infty\left|\frac{\sin(na)}{n^2}\right|$$ with the $2$-harmonic series $$\sum_{n=1}^\infty\frac{1}{n^2}.$$ The latter converges, so the former is absolutely convergent, and so convergent.

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