Uniform convergence of a series of functions with cos(x) I have to solve the following problem about series of functions:
Study the uniform convergence of the following series : $$\sum_{n=1}^\infty \frac{1}{\sqrt{n}}\left(1-\cos\left(\frac{x}{\sqrt{n}}\right)\right) $$
I studied the pointwise convergence first, and I got that the series converges pointwise for every x Real.
I tried to use the M-test, but it fails because I get a sequence $M_n$ such that $\sum_{n=0}^\infty M_N$ diverges. The fact that the M-test fails doesn't mean that the series doesn't converge uniformly right? If that's the case, how can I prove that it does or doesn't uniformly converges in R?
Thanks
 A: The series converges uniformly on any bounded interval.   Note that $\cos y \geqslant 1 - \frac{y^2}{2}$ for all $y \in \mathbb{R}$, and, thus, $\frac{1}{\sqrt{n}} \left(1 - \cos \frac{x}{\sqrt{n}} \right) \leqslant \frac{x^2}{2n\sqrt{n}}$.  Uniform convergence on bounded intervals follows by the Weierstrass M-test.
However, convergence is not uniform for all $x \in \mathbb{R}$. Taking $x_n = \frac{\pi \sqrt{2n}}{2}$, we have for all $n < k \leqslant 2n$,
$$\frac{\pi}{4} \leqslant \frac{x_n}{2 \sqrt{k}} <  \frac{\sqrt{2}\pi}{4}<\frac{\pi}{2},$$
and
$$1 = 2\cdot \sin^2 \frac{\pi}{4} \leqslant 1 - \cos \frac{x_n}{\sqrt{k}}= 2\sin^2 \frac{x_n}{2\sqrt{k}} < 2\cdot \sin ^2 \frac{\pi}{2} = 2$$
Thus,
$$\sup_{x \in \mathbb{R}}\left|\sum_{k = n+1}^\infty \frac{1}{\sqrt{k}}\left(1 - \cos \frac{x_n}{\sqrt{k}} \right) \right|\geqslant \\\sum_{k=n+1}^{2n}\frac{1}{\sqrt{k}}\left(1 - \cos \frac{x_n}{\sqrt{k}} \right)> 1\cdot\sum_{k = n+1}^{2n}\frac{1}{\sqrt{k}}> \frac{n}{\sqrt{2n}}\underset{n \to \infty}\longrightarrow +\infty,$$
proving non-uniform convergence on $\mathbb{R}$. (Uniform convergence requires the limit of the LHS as $n \to \infty$ to be zero).
