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> Suppose $\lim\_{n \to \infty}a\_n=a$. Prove that $$\lim\_{n\to \infty} \frac{1}{2^n} \sum\_{k=0}^n \binom{n}{k}a\_k=a\,.$$
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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?