Prove that $\sum_{n = 1}^\infty \frac{(-1)^n}{\sqrt{n}} \sin(1+\frac{x}{n})$ I know I can't use Weierstrass M-test for uniform convergence because it's not all positive elements. What other criteria can I use? It is the $(-1)^n$ that gives me conflict.
Sorry if I ask obvious questions, but I'm learning everything from books, and the books write that this is obvious.
 A: Note that $\sin \left(1 + \frac{x}{n}\right) = \sin(1)\cos \left(\frac{x}{n}\right)+ \cos(1)\sin \left(\frac{x}{n}\right),$ and
$$\tag{*}\sum_{n=1}^m\frac{(-1)^n}{\sqrt{n}}\sin \left(1 + \frac{x}{n}\right) = \sin(1)\sum_{n=1}^m\frac{(-1)^n}{\sqrt{n}}\cos \left(\frac{x}{n}\right)+ \cos(1)\sum_{n=1}^m\frac{(-1)^n}{\sqrt{n}}\sin \left(\frac{x}{n}\right)$$
For the second sum on the RHS of (*), we have uniform convergence on any compact subset $D$ (where $|x| \leqslant M$) by the Weierstrass M-test since
$$\left|\frac{(-1)^n}{\sqrt{n}}\sin \left(\frac{x}{n}\right) \right| = \frac{\left|\sin \left(\frac{x}{n}\right) \right| }{\sqrt{n}} \leqslant\frac{\left|\frac{x}{n}\right| }{\sqrt{n}} \leqslant \frac{M}{n^{3/2}}  $$
For the first sum on the RHS of (*), note that  $\sum \frac{(-1)^n}{\sqrt{n}}$is a convergent alternating series and is uniformly convergent as there is no $x$-dependence. Furthermore the sequence $\cos\left(\frac{x}{n} \right)$ is uniformly bounded and eventually monotone since for all $n > \frac{2M}{\pi}$ we have
$$- \frac{\pi}{2}< -\frac{M}{n} \leqslant \frac{x}{n} \leqslant \frac{M}{n} < \frac{\pi}{2},$$
so that $\cos \left(\frac{x}{n}\right)$ is monotone increasing as $n \to \infty$. By Abels' test the sum converges uniformly on $D$.
