Find sets on which $\sum^{\infty}_{n=1} \frac{\cos(nx)}{e^{nx}}$ is uniformly convergent. I need to find $x \in \mathbb{R}$ such that:
$$\sum^{\infty}_{n=1} \frac{\cos(nx)}{e^{nx}}$$
is pointwise convergent and characterise sets on which the series is uniformly convergent.
Now, I think that the series is pointwise convergent for every $x > 0$. Because in that case:
$$\sum^{\infty}_{n=1} \frac{\cos(nx)}{e^{nx}} \leq \sum^{\infty}_{n=1} \frac{1}{e^{nx}} \leq \sum^{\infty}_{n=1} \frac{1}{n^2} \rightarrow \text{converges}$$
If $x = 0$ then we get:
$$\sum^{\infty}_{n=1} \frac{\cos(nx)}{e^{nx}} = \sum^{\infty}_{n=1} 1 = \infty$$
if $x < 0$:
$$\sum^{\infty}_{n=1} \frac{\cos(nx)}{e^{-nx}} = \sum^{\infty}_{n=1} e^{nx}\cos(nx) = \infty$$
I don't know how to find sets on which the series is uniformly convergent.
 A: Your comparison test above for pointwise convergence is not valid since the cosine term changes sign.  But you can easily fix this as an absolute convergence comparison test, since $\left|\frac{\cos nx}{e^{nx}} \right| \leqslant  e^{-nx}$ and for $x > 0$ the series $\sum e^{-nx}$ is a convergent geometric series.
If $x \in [a,\infty)$ where $a > 0$, then $\displaystyle\left|\frac{\cos n x}{e^{nx}} \right| \leqslant e^{-na}$ and the series converges uniformly by the Weierstrass M-test on $[a,\infty)$.
Obviously, the series diverges for any $x \leqslant 0$ as you showed.
Finally we can show that the series fails to converge uniformly on $(0,\infty)$.
Note that for any $n \in \mathbb{N}$ and $x_n = \pi/(6n)\in (0,\infty)$, we have $1/2 < \cos kx_n \leqslant \sqrt{3}/2$ for $n < k \leqslant 2n$, and
$$\left|\sum_{k=n+1}^{2n}\frac{\cos kx_n}{e^{kx_n}}\right| = \sum_{k=n+1}^{2n}\frac{\cos kx_n}{e^{kx_n}} > n \cdot \frac{1/2}{e^{2nx_n}} = \frac{n}{2e^{\pi/3}} \underset{n \to \infty}\longrightarrow\infty,$$
and the series fails to converge uniformly on $(0,\infty)$ by violation of the uniform Cauchy criterion.
