Suppose that for all $n\in\mathbb{N},$ $(a_{m,n})_m$ is a sequence of positive real numbers such that $$\sum_{m=0}^\infty a_{m,n}=C<\infty,$$ and that $(a_{m})_m$ is another sequence of positive real numbers such that $$\sum_{m=0}^\infty a_{m}=C.$$
Suppose also that for all $m\in\mathbb{N}$, $a_{m,n}\to a_m.$ Is it true that
$$\sum_{m=0}^\infty a_{m,n}e^{itm}\to\sum_{m=0}^\infty a_{m}e^{itm},$$ for all $t\in\mathbb{R}?$
On
Yes, the result holds. Sketch: In fact as $n\to \infty,$
$$\tag 1 \sum_{m=1}^{\infty}|a_{mn}-a_m| \to 0,$$
which implies the result.
Elementary approach: To prove $(1),$ start with this: Given $\epsilon>0,$ there exists $M$ such that $\sum_{m=1}^{M}a_m >C-\epsilon;$ argue that therefore $\sum_{m=1}^{M}a_{mn} >C-2\epsilon$ for large $n.$
Measure theory approach: Suppose $f_1,f_2,\dots \in L^1$ and $f_n \to f$ a.e. If $\int |f_n|\to \int |f|,$ then $\int|f_n-f| \to 0.$ This result is well known and has appeared on MSE a zillion times; it involves a slightly clever use of Fatou's lemma. In your problem we would be working in $l^1(\mathbb N).$
If you add another constraint: $a_{m,n}\le a_m$ for sufficiently large $m,n$, then the limit you hypothesized is true.
Let us assume $\lim_{n\to\infty}a_{m,n}=a_m$.
Then, $$\begin{align} \lim_n\sum^\infty_{m=0}a_{m,n}e^{itm} &=\lim_n\int a_{m,n}e^{itm}d\mu \qquad{(1)}\\ &=\int \lim_n a_{m,n}e^{itm}d\mu \qquad{(2)}\\ &=\int a_m e^{itm} d\mu \\ &=\sum^\infty_{m=0}a_m e^{itm} \end{align}$$
$(1):$ writing a sum as a Lebesgue integral with respect to the counting measure on $\mathbb N$.
$(2):$ exchanging limit and integration as permitted by dominated convergence theorem, since $|a_{m,n}e^{itm}|\le a_m$ and $\int a_m d\mu$ comverges.