Convergence of $\sum_{n=1}^\infty \frac{\sin nx }{n}$: why can't we use the M test? 
Why can't we use the  M test for testing the convergence of this series:  $$\sum_{n=1}^\infty \frac{\sin nx }{n}?$$ 


We have $\frac{\lvert \sin nx \rvert}{n}\leq \frac{1}{n}$ hence the series must be divergent.
 A: The $M$-test would go exactly the other way ("thing $\leq$ convergent $\Rightarrow$ convergent"): you need to upper bound the summand by a quantity $M_n$ (independent of $x$) such that $\sum_n M_n$ is absolutely convergent. (To conclude the series of functions is itself uniformly convergent.)
Here, you upper bound it by something that is divergent. So the M-test does not tell you anything -- you cannot conclude anything from this. ("thing $\leq$ divergent $\Rightarrow$ nothing")
As an illustration: consider
$$
\sum_{n=1}^\infty \underbrace{\frac{\sin n x}{ n^2}}_{f_n(x)}
$$
which is a series of functions absolutely convergent. But you can say exactly the same, namely $\lvert f_n(x)\rvert \leq \frac{1}{n}$ for all $x\in\mathbb{R}
$... clearly, since does not imply divergence of $\sum_{n=1}^\infty f_n(x)$.
A: I don't know about the Wierstrass M test, but on the contrary, I have used Squeeze theorem.
$-1 \le \sin {nx} \le 1$
$-\frac{1}{n} \le \sin {nx}{n} \le \frac{1}{n}$
$\lim_{n \rightarrow \infty} \frac{1}{n}=\lim_{n \rightarrow \infty} -\frac{1}{n}=0$
Hence by squeeze theorem , $\sum _{n=1}^\infty \frac{\sin{nx}}{n}$ is also convergent since limit is defined.  
