QUESTION: Let $$ g_n(x) =\begin{cases} 1-\cos(2n\pi x) & \mbox{if} \; 0\leq x \leq 1 \\ 0 & \text{otherwise}\\ \end{cases} $$ be the density function of $G_n$. (I've already checked that it is indeed a density function). However, I'm struggling to show that $G_n$ converges weakly to the uniform distribution on $[0,1]$ and also to show that its densities $g_n$ do not converge.
MY ATTEMPT: The hint is to use convergence in distribution (by definition). So, I've tried to use Taylor's series, otherwise, I'm not sure whether this is the right way... I guess I'm not providing a precise proof. In addition I don't know how to show that the sequence $g_n$ doesn't converge.
COMMENT: I did not understand why the downvote? Is there something that someone can suggest as a improvement? So than, I'll edit.
If $x \in (-\infty, 0]$, then $$G_n(x) = \int_{-\infty}^{x} g_n(y) \ \mathrm{d} y= 0$$
And if $x \in [0, 1]$, it's clear that $$G_n(x) = x - \frac{\sin(2\pi n x)}{2 \pi n}$$
Finally if $x > 1$ it's also easy to see that $G_n(x) = 1$. Putting it all together it's clear that $G_n(x) \to G(x)$ for every $x \in \mathbb{R}$, where $G$ is the cumulative distribution function of a continuous uniform random variable with parameters $0$ and $1$.