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?

$$w_n-z=\sum_{k=0}^n2^{-n}\binom{n}k(z_k-z)$$
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 $n\ge r$.
