Sequence Proof with Binomial Coefficient Suppose  $\lim_{n\to \infty }z_n=z$. 
Let  $w_n=\sum_{k=0}^n {2^{-n}{n \choose k}z_k}$
Prove $\lim_{n\to \infty }w_n=z$.
I'm pretty sure I need to use $\sum_{k=0}^\infty{n \choose k}$ = $2^{n}$ in the proof. Help? Thoughts?
 A: It's probably not what you are looking for, but here is a solution using probability. 
Let $(X_j)$ be i.i.d. Bernoulli (1/2) random variables, 
and let $S(n)=X_1+\cdots+X_n$. The law of large numbers guarantees that $S(n)\to \infty$ almost surely, so by bounded 
convergence we have 
$$w_n=\mathbb{E}(z_{S(n)})\to \mathbb{E}(z_{\infty})=z.$$

Hint: Since $\sum_{k=0}^\infty 2^{-n}{n \choose k}=1$, the result is true for constant sequences. So you may subtract off $z$, i.e., suppose without loss of generality that $z=0$. When $z_n\to 0$, it  isn't too hard to bound $|w_n|$.
A: $$w_n-z=\sum_{k=0}^n2^{-n}\binom{n}k(z_k-z)$$
Fix $\epsilon>0$. There is an $m\in\Bbb Z^+$ such that $|z_k-z|<\frac{\epsilon}2$ whenever $k\ge m$. Clearly
$$\lim_{n\to\infty}\sum_{k=0}^m2^{-n}\binom{n}k=0\;,$$
so there is an $r\ge m$ such that $$\sum_{k=0}^m2^{-n}\binom{n}k|z_k-z|<\frac{\epsilon}2$$ whenever $n\ge r$.
Then
$$\begin{align*}
|w_n-z|&\le\sum_{k=0}^m2^{-n}\binom{n}k|z_k-z|+\sum_{k=m+1}^n2^{-n}\binom{n}k|z_k-z|\\
&\le\sum_{k=0}^m2^{-n}\binom{n}k|z_k-z|+\sum_{k=0}^n2^{-n}\binom{n}k|z_k-z|\\
&\le\frac{\epsilon}2+\frac{\epsilon}2\sum_{k=0}^n2^{-n}\binom{n}k\\
&=\epsilon
\end{align*}$$
for $n\ge r$.
