# 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?

• Continuity yes, differentiability no. The uniform convergence of the derivatives gives you differentiability. – Daniel Fischer Jun 6 '15 at 14:56
• In fact, the $n$-th derivatives all converge uniformly for any $n$, so the limit is smooth. – JHance Jun 6 '15 at 15:05

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.