A well-known theorem states that if $\sum_n \frac{d}{dx} a_n(x)$ is uniformly convergent then the series $\sum_n a_n $ can be differentiated term by term, i.e. $\frac{d}{dx}\sum_n a_n(x) = \sum_n \frac{d}{dx} a_n(x)$.
Is the converse of this theorem true? That is, if a series of functions is not uniformly convergent does that imply that you cannot differentiate term by term? My intuition says no but I am also wondering if the failure of uniform convergence implies anything of interest with regard to derivatives.
No. The series $\sum_{n=1}^\infty x^n$ does not converge uniformly on $[0,1)$ to $\frac1{1-x}$. Nevertheless, it is true that$$\bigl(\forall x\in[0,1)\bigr):\sum_{n=1}^\infty nx^{n-1}=\frac1{(1-x)^2}=\left(\frac1{1-x}\right)'.$$