Suppose $\lim_{n \to \infty}a_n=a$, Prove $\lim_{n\to \infty} \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k}a_k=a$.

Question

Suppose $\lim_{n \to \infty}a_n=a$. Prove that $$\lim_{n\to \infty} \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k}a_k=a\,.$$

Here is a proof,

Proof: Consider $t_{n,k}=\frac{1}{2^n}\binom{n}{k}$,$0 \leq t_{n,k}= \frac{n(n-1)...(n-k+1)}{2^nk(k-1)...1}\leq\frac{n^k}{2^n} $ and by squeeze theorem, $\lim_{n \to \infty}t_{n,k}=0$. Also, $\sum_{k=0}^n \frac{1}{2^n}\binom{n}{k}=1$, so then by Silverman–Toeplitz theorem, we have $\lim_{n\to \infty} \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k}a_k=a$.

Is this proof correct? Is there other cool proofs?

Answer 1

Say $|a_n| \le C$ for each $n$. For $\epsilon > 0$, take $N$ s.t. $|a_n-a| < \epsilon$ for $n \ge N$. Then for $M \ge N^2$, $$\frac{1}{2^M}\sum_{k=0}^M {M \choose k}|a_k-a| \le \frac{1}{2^M}\sum_{k=0}^\sqrt{M} {M \choose k}2C+\frac{1}{2^M}\sum_{k=\sqrt{M}}^M {M \choose k}\epsilon.$$ So, $\limsup_{M \to \infty} |\frac{1}{2^M}\sum_{k=0}^M {M \choose k} a_k - a| \le \epsilon$. This holds for each $\epsilon > 0$.

Answer 2

There's a pretty standard exercise in which we are asked to show that if $a_k\rightarrow a$ then $$\biggl(\frac{1}{n}\sum_{k=1}^n a_k\biggr)\rightarrow a\,.$$ It's reasonable to suspect that a similar approach there might work here.

After one starts down this road, you eventually realize that you'll need to show that

$$\lim_{n\rightarrow\infty}\frac{1}{2^n}\sum_{k=0}^N\binom{n}{k}=0$$

where $N$ is some possibly-big, but-nevertheless-fixed positive integer. To accomplish this, we notice $$\sum_{k=0}^N\binom{n}{k}\leq\sum_{k=0}^N\frac{n^k}{k!}=\sum_{k=0}^N\frac{N^k}{k!}\left(\frac{n}{N}\right)^k\leq\left(\frac{n}{N}\right)^Ne^N$$ at least once $n$ surpasses $N$. This inequality gives the limit we wanted.

Now we are ready to start the formal presentation. Let $\epsilon>0$. Choose $N_1$ in $\mathbb{N}$ such that $$|a_n-a|<\frac{\epsilon}{2}$$ for every $n\geq N_1$. Let $M$ satisfy the inequality below for every $n$ in $\mathbb{N}$ $$|a_n-a|<M\,.$$ Choose $N_2$ such that $$\frac{1}{2^n}\sum_{k=0}^{N_1-1}\binom{n}{k}<\frac{\epsilon}{2(M+1)}$$ for every $n\geq N_2$. We now observe the scratch-paper inequalities that inspired the approach:

\begin{align*}\left|\biggl(\,\sum_{k=0}^n\frac{1}{2^n}\binom{n}{k}a_k\biggr)-a\,\right| &=\left|\sum_{k=0}^n\frac{1}{2^n}\binom{n}{k}(a_k-a)\right|\\ &\leq \frac{1}{2^n}\sum_{k=0}^{N_1-1}\binom{n}{k}|a_k-a|+\sum_{k=N_1}^n\frac{1}{2^n}\binom{n}{k}|a_k-a| \end{align*} for sufficiently large $n$. In particular, if we take $n>\max(N_1, N_2)$, we then get $$\left|\biggl(\,\sum_{k=0}^n\frac{1}{2^n}\binom{n}{k}a_k\biggr)-a\,\right|<\epsilon\,.$$