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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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    This is the Clausen function. It is closely related to the dilogarithm.2012-06-18

5 Answers 5

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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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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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    @PeterTamaroff Indeed. I was referring to the "Mr." part, but as Brian has told us, it's all the same to him =).2012-06-17
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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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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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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 $a$.