If $\mathbf {i)} f_n\in C^1(\overline U)$ ($U$ is convex), $\mathbf {ii)} \sum\limits_{n\ge1}df_n$ converges unifromly on $U$ and $\mathbf {iii)} \sum\limits_{n\ge1}f_n$ converges in an arbitrary point $x_0\in U$ then,
$\mathbf {1)}$ $\sum\limits_{n\ge1}f_n$ converges uniformly on $\overline U$
$\mathbf {2)}$ $f\in C^1(\overline U)$
$\mathbf {3)}$ $df=\sum\limits_{n\ge1}df_n$
First, $S_N=\sum\limits_{n\le N}f_n$ is $C^1(\overline U)$ then we have,
$S_N(x_0+vh)=S_N(x_0)+\displaystyle\int_0^h\partial_vS_N(x_0+sv)ds\tag1$
and $\partial_vS_N\to\sum\limits_{n\ge1}\partial_vf_n$ uniformly, so the integral on the RHS of $(1)$ exists
(if it were not uniformly, would the integral also exist ?)
since $S_N(x_0)\to f(x_0)$ the whole expression on the RHS of $(1)$ exists.
so what did we reach now ?