# Determine if the following series is convergent or divergent $\sum_{n=1}^{\infty}\frac{ \sin{nx}}{n}$

I've been doing exercises with series for some time and now I've got an exercise that I'm supposed to solve using Abel or Dirichlet criteria. The problem is :

Determine if the following series is convergent or divergent by using Abel or Dirichlet criteria:

$$\sum_{n=1}^{\infty}\frac{ \sin{nx}}{n}$$

I do not know what to do, because:

• First of all, Abel criteria says that if $\sum_{n=1}^{\infty}{x_n}$ is convergent and $y_n$ is a monotonous and bounded sequence, then $\sum_{n=1}^{\infty}{x_n y_n}$ is a convergent series, so in my case $\sin(nx)$ is bounded and monotonous, but $\sum_{n=1}^{\infty}{1 \over n}$ isn't convergent, so I think I can't use Abel cr.

• Second, Dirichlet criteria says that the series $\sum_{n=1}^{\infty}{x_n}$ should have the sequence of partial sums bounded and $y_n$ should be decreasing and convergent to zero for $\sum_{n=1}^{\infty}{x_n y_n}$ to be convergent. In my case, whatever $\sum_{n=1}^{\infty}{x_n }$ I take it will not have bounded sequence of partial sums, so what should I do ? What should I use ? May be I am wrong somewhere ?

Thanks.

• no, the question is : diverges it or not ? – Ivan Gandacov Jan 20 '15 at 20:16
• I want to know what should I use , Dirichlet or Abel, and I listed my misunderstandings. – Ivan Gandacov Jan 20 '15 at 20:17

Dirichlet's test is appropriate here. Since $\dfrac 1n \to 0$, to show that the series converges it suffices to show that there is a constant $M$ (depending on $x$) with the property that $$\left| \sum_{k=1}^n \sin kx \right| \le M$$ for all $n \ge 1$. This can be done with elementary trigonometry.
Edited: if $x$ is a multiple of $\pi$ then $\displaystyle \sum_{k=1}^n \sin kx = 0$ for all $n$. Otherwise use the identities $$\cos(k + \frac 12)x = \cos kx \cos \frac x2 - \sin kx \sin \frac x2$$ and $$\cos(k - \frac 12)x = \cos kx \cos \frac x2 + \sin kx \sin \frac x2$$ to get $$\left[ \cos(k - \frac 12)x - \cos(k + \frac 12)x \right] = 2 \sin kx \sin \frac x2.$$ It follows that $$\cos \frac x2 - \cos (n+\frac 12)x = \sum_{k=1}^n \left[ \cos(k - \frac 12)x - \cos(k + \frac 12)x \right] = 2 \sin \frac x2 \sum_{k=1}^n \sin kx$$ so that $$\sum_{k=1}^n \sin kx = \dfrac{\cos \frac x2 - \cos (n+\frac 12)x}{2 \sin \frac x2}.$$ Since the cosine function is bounded by $1$ and $-1$ you have the inequality $$\left| \sum_{k=1}^n \sin kx \right| \le \frac{1}{|\sin \frac x2|}$$ for all $n$. This bound is finite since $x$ is not a multiple of $\pi$.
You can use Dirichlet criteria because $\sum_{k=1}^{n}\ \sin{kx}$ is bounded.
To see this, notice that $\sum_{k=1}^{n}\ \sin{kx}$ = $Im(\sum_{k=1}^{n}\ e^{ikx})$.