TITLE: If $a_n \to z$, then does $\frac{{n \choose 1} a_1 + {n \choose 2} a_2 \dots {n \choose n} a_n}{2^n} \to z$?
QUESTION [7 upvotes]: 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 $n\ \to \infty$?

REPLY [2 votes]: 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 
$$\lim_{n\to\infty}\frac{1}{2^n}\sum_{i=1}^n \binom{n}{i}a_i=z.$$