Prove : $\sum_{n=1}^{\infty}\left(\frac{x}{2}\right)^{n}\sin\left(\frac{\pi x}{4}\right)$ is not uniformly convergent in $\left({-2,2}\right)$

I've been having a hard time proving that the sequence $$\sum_{n=1}^{\infty}\left(\frac{x}{2}\right)^{n}\sin\left(\frac{\pi x}{4}\right)$$
Does not uniformly converge for $x$ in the interval $\left({-2,2}\right)$.

I have concluded it might not be uniformly convergent due to its similarily to $x^n$ and considering a $\sin$ appears in the product, but using the Cauchy test for uniformly convergence was not helpful,

Because $\forall{N}\in\mathbb{N},\exists{m>n>N}:$ $$|\sum_{k=n}^{m}\left(\frac{x}{2}\right)^{k}\sin\left(\frac{\pi x}{4}\right)|$$

Then for any $x\in\left({-2,2}\right)$, we'd get $0\leq|\frac{x}{2}|<1$, and therefore cannot be assesed to find an appropriate $\varepsilon_0>0$.

How would you go about proving uniform disconvergence in this case?

Thank you in advance!

Solution 1:

For the series to converge uniformy it is necessary that $\sup_x |\left(\frac{x}{2}\right)^{n}\sin\left(\frac{\pi x}{4}\right)| \to 0$.

But $\sup_x |\left(\frac{x}{2}\right)^{n}\sin\left(\frac{\pi x}{4}\right)\geq |(\frac {(2-\frac 1 n)} 2 )^{n}\sin \frac {\pi (2-\frac 1 n)} 4|$ and this last quantity does not tend to $0$. In fact, it tends to $e^{-1/2}$. Hence, the series is not uniformly convergent.