Uniform convergence implies continuity and differentiability? For example: Suppose I have the following series:
$$\sum_{k=0}^{\infty}e^{-k}\sin(kt)$$
The Weierstrass-M-Test shows that the series is uniformly convergent on $\mathbb R$. Does this imply differentiability and continuity on $\mathbb R$ aswell?

Yes, since each finite sum is continuous, then the uniform convergence of the series implies the continuity of the limit on any compact of $\mathbb{R}$, thus the continuity of the limit sum on $\mathbb{R}$.
For the differentiability, you can check that the series
$$
\left|\sum_{k=0}^{\infty}k\:e^{-k}\cos(kt)\right|\leq\left|\sum_{k=0}^{\infty}k\:e^{-k}\right|=\frac{e}{(1-e)^2}<+\infty
$$ is (normally) uniformly convergent on $\mathbb{R}$, giving the desired the differentiability of the limit sum.
