Let $f_n(x)$ be concave functions.
I know that \begin{align} g(x)=\sum_{n=1}^N a_nf_n(x) \end{align} where $a_n \ge 0$ and $N<\infty$ is concave.
My question is if the sum becomes infinite sume is $g(x)$ still concave? To me it seems yes, since in the proof we do not assume anyting about finitness of the sum:
\begin{align} g((1-t)x+ty)&=\sum a_nf_n((1-t)x+ty) \ge \sum a_n ( (1-t)f_n(x)+tf_n(y))\\ &=(1-t) g(x)+tg(y) \end{align}
As you may or may not have guessed from the comments, if you know that the sum converges (pointwise) for every argument you want to check it for (that is, in this case, for $(1-t)x+ty, (1-t)x$, and $ty$ then the conclusion is valid.
The proof of the inequality would usually be done by reducing it to a finite sum and some $\varepsilon, \delta $ logic, but it's rather straightforward.