If $a_n \to z$, then does $\frac{{n \choose 1} a_1 + {n \choose 2} a_2 \dots {n \choose n} a_n}{2^n} \to z$? It is well known that $\sum_{k=0}^n{n\choose k} =2^n$.
My question: If $z$ is the limit point of an infinite sequence of real numbers $\{ a_n \}$, then does $$\frac{{n \choose 1} a_1 + {n \choose 2} a_2+ \cdots+ {n \choose n} a_n}{2^n}$$ converge to $z$ as $n\ \to \infty$?
 A: Something far more general is true. Define the doubly half-infinite matrix $ c_{mn}$,
$$ c_{mn} = \frac1{2^m}\binom{m}{n}\mathbb 1_{n\le m} $$
then you're asking if
$$ t_m := \sum_{n=0}^\infty c_{mn} a_n  \to z?$$
According to Hardy's Divergent Series, Theorem 2 (page 43) (apparently due to Toeplitz and Schur):

Theorem 2. (Paraphrased) For any such infinite matrix $(c_{mn})_{m,n\ge 0}$, the statement
  $$\text{“$a_n \to z$ implies $t_m \to z$''}$$
  is true iff the following conditions for $c_{mn}$ are satisfied:
  
  
*
  
*$\sum_{n=0}^\infty |c_{mn} | < H $ where $H<\infty$ is uniform in $m$,
  
*for each $n$, $c_{mn} \xrightarrow[m\to\infty]{} 0$,
  
*$\sum_{n=0}^\infty c_{mn} \xrightarrow[m\to\infty]{} 1$.
  

For your specific $c_{mn}$, for each of these conditions:


*

*$\sum_{n=0}^\infty |c_{mn} | =\sum_{n=0}^\infty c_{mn}    = 2^{-m}\sum_{n=0}^m \binom{m}{n} = 1$. 

*$c_{mn}  = \frac{m!}{2^m n!(m-n)!} \le \frac{m^n}{n!2^m} \xrightarrow[m\to\infty]{} 0.$

*see 1.


So the assumptions for Theorem 2 are verified, and the result holds.
(This type of result is a "regularity" result for a summation method, and such methods are the main focus of Hardy's book.)
A: Below is not entirely rigorous.
Since $\{a_n\}$ is convergent, we can pick $N$ such that $a_m$ is close to $z$ for any $m\ge N$.
Let $n$ large, and due to the fact that if $N=O(1)$ (implies $o(n)$),
 sum of $\sum_{k=1}^N {{n}\choose{k}}=O(n^N)$
We have $\frac{{n \choose 1} a_1 + {n \choose 2} a_2+ \cdots+ {n \choose n} a_n}{2^n} $ is close to $\frac{{n \choose 1} (z-\epsilon) + {n \choose 2} (z-\epsilon)+ \cdots+ {n \choose n} (z-\epsilon)}{2^n} $,
close to $z$.
A: Let $\epsilon>0$ and define $b_n=a_n-z$. Since $a_n\to z$ there exists $N$ such that for all $n\geq N$, $|b_n|<\epsilon$. Now, consider the limit
$$\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}a_i=\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}(b_i+z)$$
$$=\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}b_i+\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}z=z+\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}b_i.$$
We may simplify the problem and only consider the remaining limit. Note that proving the original statement is equivalent to proving 
$$\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}b_i=0.$$
For $n\geq N$ split the sum into two parts, $1\leq i<N$ and $i\geq N$. We have
$$\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}b_i=\lim_{n\to\infty}\left(\frac{1}{2^n}\sum_{i=1}^{N-1} \binom{n}{i}b_i+\frac{1}{2^n}\sum_{i=N}^{n} \binom{n}{i}b_i\right)$$
$$=\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^{N-1} \binom{n}{i}b_i+\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=N}^{n} \binom{n}{i}b_i.$$
Now, the first limit is clearly zero as $\binom{n}{i}=O(n^i)$ which implies the numerator is some polynomial in $n$ while the denominator is an exponential in $n$. Thus, our problem is distilled down to showing 
$$\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=N}^{n} \binom{n}{i}b_i=0.$$
To do this, note that
$$\frac{1}{2^n}\sum_{i=N}^{n} \binom{n}{i}b_i\leq \frac{1}{2^n}\sum_{i=N}^{n} \binom{n}{i}|b_i|<\frac{1}{2^n}\sum_{i=N}^{n} \binom{n}{i}\epsilon\leq \frac{\epsilon}{2^n}\sum_{i=1}^{n} \binom{n}{i}=\epsilon.$$
Thus,
$$\frac{1}{2^n}\sum_{i=N}^{n} \binom{n}{i}b_i$$
is bounded by every positive number and hence is $0$ in the limit. We conclude 
$$\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}a_i=z.$$
