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

Question

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

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We have $\frac{\lvert \sin nx \rvert}{n}\leq \frac{1}{n}$ hence the series must be divergent.

Answer 1

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 $\sum_{n=1}^\infty f_n(x)$.

Answer 2

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.