Why series $\sum_{n=1}^{\infty} \frac{e^{-\vert{x}\vert}}{n^3}$ is uniformly convergent in $x \in (-\pi,\pi)$?
My answer: ${{e^{-\vert{x}\vert}} \le1 }$ for $x \in (-\pi,\pi)$ implies $\frac{e^{-\vert{x}\vert}}{n^3} \le$ $\frac{1}{n^3}$, which is convergent and hence by Weierstrass $M$-test given series is uniformly convergent.
Please correct if I am wrong.
Yes, you are correct. Since $e^{-|x|}\leq 1$ for all real number $x$, it follows that the series $$\sum_{n=1}^{\infty} \frac{e^{-\vert{x}\vert}}{n^3}$$ is uniformly convergent in $\mathbb{R}$.