A limit involving binomial coefficients and square roots I am trying to evaluate the following limit
$$\lim_{n\to\infty}\frac{1}{2^n\sqrt{n}}\sum^n_{k=1}\binom{n}{k}\sqrt{k},$$
but to no avail. I experimented with some large values of $n,$ and it seems like the limit is $\frac{1}{\sqrt{2}}.$ In fact, I am able to prove that $$\frac{1}{2^n\sqrt{n}}\sum^n_{k=1}\binom{n}{k}\sqrt{k}\leq\frac{1}{\sqrt{2}}$$
using the fact that the square root function is concave, as follows. We show that $$\sum^n_{k=0}\binom{n}{k}\sqrt{k}\leq \sum^n_{k=0}\binom{n}{k}\sqrt{\frac{n}{2}}=2^n\sqrt{\frac{n}{2}}.$$
Indeed, for $k\leq\frac{n}{2},$ one has $\binom{n}{k}\sqrt{k}\leq\binom{n}{k}\sqrt{\frac{n}{2}}.$ But since square root is concave, we get $\sqrt{\frac{n}{2}}-\sqrt{k}\geq\sqrt{n-k}-\sqrt{\frac{n}{2}}.$ Moreover, $\binom{n}{k}=\binom{n}{n-k},$ so it follows that $$\sum^n_{k=0}\binom{n}{k}\left(\sqrt{k}-\sqrt{\frac{n}{2}}\right)\leq 0,$$
as desired. With this, I suspect that one may obtain a lower bound of the sum and apply squeeze theorem to obtain the final limit. Any ideas on this?
 A: From the suggestion of @Jean Marie, you can solve the problem as follows in a more general way, as explained in @Sayan's answer here.
First, let $f(n)=\sum_{k=0}^n\binom{n}{k}p^k(1-p)^{n-k}g\left(\frac{k}{n}\right)$ for $0\leq p\leq1$ and some function $g$ of interest. In your case, $p=1/2$ and $g(x)=\sqrt{x}$.
You can view $f(n)$ as the expected value of a Binomial random variable $Y\sim\mathrm{Bin}(n, p)$, so that $f(n)=\mathbb{E}[g(Y/n)]$. This can be useful if $g$ is concave (convex) and you want to find an upper (lower) bound since you can use Jensen's inequality: $\mathbb{E}[g(Y/n)]\overset{(\geq)}{\leq} g(\mathbb{E}[Y/n])=g(p)$.
However, to determine the limit, one needs to use the Strong Law of Large Numbers (SLLN) and apply it to Bernoulli variables. So let $X_1, \dots, X_n$ be i.i.d Bernoulli random variables with success probability $p$. By the SLLN, one can write
$$\bar{X_n}:=\frac1n\sum_{i=1}^n X_i \overset{a.s.}{\to} \mathbb{E}[X_1]=p,$$
and more generally for continuous functions $g$, we have $g(\bar{X_n})\overset{a.s.}{\to}g(p)$. Suppose moreover that $g$ is such that there exist an unsigned absolutely integrable random variable $Y$ such that $|g(\bar{X_n})|\leq Y$. Then, by the Dominated Convergence Theorem,
$$\lim_{n\to \infty}\mathbb{E}[g(\bar{X_n})]=g(p).$$
Finally, using the above and the fact that $Y=n\bar{X_n}\sim \mathrm{Bin}(n, p)$, this yield
$$g(p) = \lim_{n\to \infty}\mathbb{E}[g(\bar{X_n})] = \lim_{n\to \infty}\mathbb{E}[g(Y/n)]=\lim_{n\to \infty}\sum_{k=0}^n\binom{n}{k}p^k(1-p)^{n-k}g\left(\frac{k}{n}\right)=\lim_{n\to \infty} f(n),$$
which is the quantity you are interested in.
In this setting, $p=1/2$ and $g(x)=\sqrt{x}$, so that $\lim_{n\to \infty} f(n) = g(p) = \frac{1}{\sqrt{2}}$ as you predicted.
