Suppose $\{f_n\}$ is a sequence of analytic in a region $D$ suppose $\{f_n\}$ is a sequence of analytic in a region $D$ such that $|f_n(z)|\leq M_n$, where $\sum_n M_n<\infty$, and $\lim_{z\to z_0}f_n(z)=L_n$.
Show that if $p=p(z)\to\infty$ as $z\to z_0$, then 
$$\lim_{z\to z_0}\{f_1(z)+f_2(z)+\dots+f_p(z)\}=\sum_n L_n.$$
Here is what I tried: 
\begin{equation*}
    \begin{aligned}
        \lim_{z\to z_0}\big(f_1(z)+f_2(z)+\dots+f_p(z)\big) & \\
        &= \lim_{z\to z_0}f_1(z)+\lim_{z\to z_0}f_2(z)+\dots+\lim_{z\to z_0}f_p(z) \\
        &= L_1+L_2+\dots+L_p \\
        &= \sum L_n
    \end{aligned}
\end{equation*}
Is this correct? To be honest, I am not entirely sure what the question is asking. Is this really just asking to show that the limit of the sums is the sum of the limits? That wouldn't really make sense in the context of the course (complex variables) but I don't see what else it could be asking.
 A: It's not correct, you didn't take into account that the number of terms depends on $z$. You can interchange summation and taking the limit for a sum with a fixed finite number of terms, but if the number of terms in the sum depends on the point, you need to argue more carefully. A simple example that shows that we cannot in general exchange summation and taking the limit when the number of terms grows to infinity is given by
$$a_k^{(n)} = \begin{cases} \frac{1}{n} &, 1 \leqslant k \leqslant n \\ 0 &, k > n. \end{cases}$$
Then we have
$$\sum_{k = 1}^n a_k^{(n)} = 1$$
for all $n$, but
$$\sum_{k = 1}^{\infty} \lim_{n\to \infty} a_k^{(n)} = 0.$$
For the case at hand, it is not important that the functions $f_n$ are analytic, or that $D\subset \mathbb{C}$ is a region. The important parts are the existence of $L_n = \lim\limits_{z\to z_0} f_n(z)$ and the majorisation $\lvert f_n(z)\rvert \leqslant M_n$ for all $z\in D$ with $\sum M_n < +\infty$.
For $\varepsilon > 0$, we can choose $N \in \mathbb{N}$ so that
$$\sum_{n = N+1}^{\infty} M_n < \frac{\varepsilon}{3}$$
since $\sum M_n < +\infty$. As $p(z) \to +\infty$ for $z\to z_0$, we can find a neighbourhood $U_N$ of $z_0$ such that $p(z) \geqslant N$ for all $z \in U_N \cap D\setminus \{z_0\}$. For each $n \leqslant N$, we can find a neighbourhood $V_n$ of $z_0$ such that $\lvert f_n(z) - L_n\rvert < \frac{\varepsilon}{3N}$ for $z \in V_n \cap D \setminus \{z_0\}$. Then $W = U_N \cap \bigcap_{n = 1}^N V_n$ is a neighbourhood of $z_0$, and for $z \in W \cap D \setminus \{z_0\}$ we have
\begin{align}
\Biggl\lvert \sum_{n = 1}^{p(z)} f_n(z) - \sum_{n = 1}^{\infty} L_n\Biggr\rvert
&\leqslant \sum_{n = 1}^{N} \lvert f_n(z) - L_n\rvert + \sum_{n = N+1}^{p(z)} \lvert f_n(z)\rvert + \sum_{n = N+1}^{\infty} \lvert L_n\rvert\\
&< \sum_{n = 1}^N \frac{\varepsilon}{3N} + \sum_{n = N+1}^{p(z)} M_n + \sum_{n = N+1}^{\infty} M_n\\
&< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3}.
\end{align}
So for all $\varepsilon > 0$ there is a neighbourhood $W$ of $z_0$ such that
$$\Biggl\lvert \sum_{n = 1}^{p(z)} f_n(z) - \sum_{n = 1}^{\infty} L_n \Biggr\rvert < \varepsilon$$
for all $z \in W\cap D \setminus \{z_0\}$, i.e.
$$\lim_{z\to z_0} \sum_{n = 1}^{p(z)} f_n(z) = \sum_{n = 1}^{\infty} L_n.$$
