Suppose $\{z_n\}_{n = 1}^\infty \subset \mathbb{C}$ is a converging sequence. I don't understand how can you prove that
$$\lim_{n \to \infty} \sum_{k=1}^n z_k\frac{ n \choose k}{2^n} = \lim_{n \to \infty} z_n$$
I think i have to use the fact that $$(a+b)^n=\sum_{k=0}^n {n \choose k}a^{n-k}b^k.$$
I think that the statement should be:
Proof. If $z_n\to a$ then $|z_n-a|\leq M$ for some $M>0$, and for $\epsilon>0$ there is $N$ such that $|z_k-a|<\epsilon$ for $k>N$. Hence, for $n>N$, since $2^n=(1+1)^n=\sum_{k=0}^n {n \choose k}$, it follows that $$\left|\sum_{k=0}^n z_k\frac{ n \choose k}{2^n}-a\right|=\left|\sum_{k=0}^n (z_k-a)\frac{ n \choose k}{2^n}\right|\leq \sum_{k=0}^n |z_k-a|\frac{ n \choose k}{2^n}\\\ \leq M\sum_{k=0}^N \frac{ n \choose k}{2^n}+ \epsilon\sum_{k=N+1}^{n} \frac{ n \choose k}{2^n}\leq M\sum_{k=0}^N \frac{ n \choose k}{2^n}+ \epsilon.$$ Now let $n\to \infty$. What may we conclude?