How to prove that $\frac1{n\cdot 2^n}\sum\limits_{k=0}^{n}k^m\binom{n}{k}\to\frac{1}{2^m}$ when $n\to\infty$ I have solve following sum
$$\sum_{k=0}^{n}k\binom{n}{k}=n2^{n-1}\Longrightarrow \dfrac{\displaystyle\sum_{k=0}^{n}k\binom{n}{k}}{n\cdot 2^n}\to\dfrac{1}{2},n\to\infty$$
$$\sum_{k=0}^{n}k^2\binom{n}{k}=n(n+1)2^{n-2}\Longrightarrow \dfrac{\displaystyle\sum_{k=0}^{n}k^2\binom{n}{k}}{n\cdot 2^n}\to\dfrac{1}{2^2},n\to\infty$$
$$\sum_{k=0}^{n}k^3\binom{n}{k}=2^{n-3}n^2(n+3)\Longrightarrow \dfrac{\displaystyle\sum_{k=0}^{n}k^3\binom{n}{k}}{n\cdot 2^n}\to\dfrac{1}{2^3},n\to\infty$$
so I conjecture the following:
$$\dfrac{\displaystyle\sum_{k=0}^{n}k^m\binom{n}{k}}{n\cdot 2^n}\to\dfrac{1}{2^m},n\to\infty$$
$$\cdots\cdots$$
so I conjecture for $m$ be positive real number also hold.

You can write your expression as 
$$
E[\bar{X}^m]n^{m-1}
$$ 
where $\bar{X}=\frac{X_1+\dots+X_n}{n}$ with independent Bernoulli random variables $X_i$. This shows that you are off by $n^{m-1}$. Indeed, if you divide by $n^{m-1}$ you can apply the strong Law of Large Numbers to conclude 
$$
E[\bar{X}^m]\to E[X_1]^m=(1/2)^m.
$$
(*) Justification of the convergence: The SLLN gives pointwise convergence of $\bar{X}^m$ to 1/2. Furthermore, $|\bar{X}|^m$ is always bounded by $1$. Therefore we can apply the dominated convergence theorem.
