TITLE: Find $ \lim\limits_{{n \to \infty}} \frac1{2^n} \sum\limits_{k=1}^n \frac1{\sqrt{k}} \binom nk$
QUESTION [17 upvotes]: Find
$$\lim_{n \to \infty} \frac {1}{2^n} \sum_{k=1}^n \frac{1}{\sqrt{k}} \binom{n}{k}.$$
First time I thought about Stirling's approximation but didn't get anything by applying it. I would also think about a Riemann Sum, but no idea how to rewrite...
The answer is $0$.

REPLY [5 votes]: The limit is part of a broad class of limits for which the classical analysis designed Toeplitz theorem. The theorem says that given an array of real numbers, $\lambda_{n,k}: 1\le k\le n,\  n\ge1 $, such that: $i) \ \lim_{n\to\infty}\lambda_{n,k}=0, \ \forall k\in \mathbb{N}$, $ii) \lim_{n\to\infty} \sum_{k=1}^n \lambda_{n,k}=1$ and $iii)$ there exists $C>0$ such that for all positive integers $n$, $\sum_{k=1}^n |\lambda_{n,k}|\le C$, then for any convergent sequence $\delta_k$, we have 
$$\lim_{n\to\infty} \sum_{k=1}^n \lambda_{n,k} \delta_k=\lim_{n\to\infty}\delta_n .$$
In your problem set $\displaystyle \lambda_{n,k}=\frac{1}{2^n}\binom{n}{k}$ and $\displaystyle \delta_k=\frac{1}{\sqrt{k}}$, and since the conditions are satisfied, we have
$$\displaystyle \lim_{n \to \infty} \frac {1}{2^n} \displaystyle \sum_{k=1}^n \frac{\displaystyle \binom{n}{k}}{\sqrt{k}}=\lim_{n\to\infty} \frac{1}{\sqrt{n}}=0,$$
which ends the little proof.