I would try to study the convergence of this series of functions:
$$ \sum_{n=1}^{\infty} \dfrac{2^n \left(2 \sin^n (\pi x) -1\right)^n}{n}. $$
The series has no positive terms so, in order to apply some criterion, i have to study the absolute convergence, i.e.
$$ \sum_{n=1}^{\infty} \dfrac{ 2^n \left\vert2 \sin^n (\pi x) -1 \right\vert^n}{n}. $$
Now, applying the root criterion, i have $$ \lim_{n\rightarrow\infty} \sqrt[n]{\dfrac{ 2^n \left\vert2 \sin^n (\pi x) -1 \right\vert^n}{n}} = 2 \lim_{n\rightarrow\infty} \left\vert2 \sin^n (\pi x) -1 \right\vert. $$
I know that, for those $ x\in\mathbb{R}$ such that the result of the limit is equal to $1$, the series converges. But i don't know how to solve this limit. Can anyone help me?
If $\sin(\pi x)$ satisfies $|\sin(\pi x)| < 1$ then the series diverges; for such $x$ and every $\epsilon > 0$ there is an $n_0$ (which depends on $x$ and $\epsilon$) such that $|2 \sin^n(\pi x)| < \epsilon$ for all $n \ge n_0$; so take $\epsilon = \frac{1}{4}$; this implies that $|\frac{2^n(2 \sin^{n}(\pi x) -1}{n}| \geq \frac{2^n(1-\epsilon)^n}{n} = \frac{2^n(3/4)^n}{n}$ $=\frac{(3/2)^n}{n}$ for all $n \geq n_0$ which implies the series diverges.
The series also diverges for the remaining $x \in \mathbb{R}$: i.e., $x$ satisfying $|\sin(\pi x)| = 1$. For $x$ satisfying $\sin(\pi x) = -1$ note that $\frac{2^n(2\sin^n(\pi x) -1)}{n}$ $= \frac{2^n(-3)}{n}$ for odd $n$ which goes to infinity ; for $x$ satisfying $\sin(\pi x) =1$ note that $\frac{2^n(2\sin^n(\pi x) -1)}{n}$ $= \frac{2^n}{n}$ which also goes to infinity.
So the series diverges for all real $x$, and therefore converges for no real $x$.
So to relate back to your attempt, the limit you are trying to take is at least 1 for all $x$, as $\sin^n(\pi x)$ goes to 0 if $x$ satisfies $|\sin(\pi x)| < 1$.