Limit of sum of periodic function
(this assumes continuity of the $f_i$, still thinking of how to remove it)
Suppose $F(x):=\sum_i f_i(x)\to C$, where $f_i(x+p_i)=f_i(x)$ for each $i$.
Now, find an sequence of positive reals $h_N$ increasing to $\infty$ which is simultaneously close to $p_i\mathbb{Z}$ for all $i$, i.e.
$$h_N=a_i^{(N)}p_i+\varepsilon_N^{(i)}, \text{where }z_N^{(i)}\in\mathbb{Z}, \varepsilon_N^{(i)}\to 0$$
as by noting Dirichlet's simultaneous approximation theorem applied to $\{\frac{1}{p_1},\frac{1}{p_2},...,\frac{1}{p_n}\}$, we can find integers $a_i^{(N)}$ and an integer $h_N\le N$ such that for each $i$:
$$\left\vert \frac{1}{p_i}-\frac{a_i^{(N)}}{h_N}\right\vert \le \frac{1}{h_N N^{1/n}}$$
Upon rearrangement, this becomes $\vert h_N - a_i^{(N)}p_i \vert \le \frac{p_i}{N^{1/n}}\to0$
Then, for any $x$:
$$\lim_{n\to\infty}F(x+h_n)=\lim_{x\to\infty}F(x)=C$$
$$\lim_{n\to\infty}F(x+h_n)=\lim_{n\to\infty}\sum_i f_i(x+h_n)=\lim_{n\to\infty}\sum_i f_i(x+\varepsilon_n^{(i)})=\sum_i f_i(x)=F(x)$$
So, $F(x)=C$ for all $x$.
Alternative Proof
Let $P(n)$ be the statement "If a sum of $n$ periodic functions has a limit $C$, then this sum is equal to $C$ for all $x$".
If $f$ is $p$-periodic and tends to $C$, then for any $\varepsilon > 0$, there exists $N$ such that $x>N\implies \vert f(x) - C \vert < \varepsilon$. But periodicity gives that this is actually true for all $x$. As this is true for any $\varepsilon > 0$, we recover that $f=C$ for all $x$. So, $P(1)$ is true.
Suppose $P(1), P(n-1)$ are true, and consider a sum of $n$ periodic functions, $F(x)=\sum_1^n f_i(x)$ with limit $C$, where in particular, $f_n$ has period $p_n$. Then $F(x+p_n)-F(x)=\sum_1^{n-1} [f_i(x+p_n)-f_i(x)]$ is a sum of $(n-1)$ periodic functions, and converges to $0$, hence is equal to $0$ by $P(n-1)$.
So, $F$ is $p_n$-periodic, and converges to $C$, and $P(1)$ tells us that it is identical to $C$ as a result, i.e. $P(n)$ is true.
Thus, by induction, $P(n)$ is true for all $n$, and any finite sum of periodic functions with a limit at $\infty$ is constant.