Does $\sum_{n=1}^{\infty} { \frac{(-1)^{\frac{n(n-1)}{2}}}{\sqrt[3]{n^2+e^x}} }$ uniformly converge in $(-\infty, \infty)$?
As to now, what I did is say that $f_n(x) = \text {our sum}$ and say that it is $\leq \frac{1}{\sqrt[3]{n^2 + e^x}} = g(x)$. Now I've been trying to prove that $g(x)$ converges but haven't been able to do so, I have tried the Ratio test, but it is inconclusive because $x$ can also be $\infty$.
Any help would be appreciated