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.

Let $(Z_i)$ be i.i.d. Bernoulli(1/2) random variables, and set $\bar Z_n={1\over n}\sum_{i=1}^n Z_i$ be the sample average. Then $\bar Z_n\to 1/2$ almost surely by the law of large numbers.
On the other hand, $X=\sum_{i=1}^n Z_i$ has a Binomial$(n,1/2)$ distribution, so 
that, $\mathbb{P}(X=k)={n\choose k}(1/2)^n$ for $0\leq k\leq n$. 
Therefore, for every real number  $m\geq 0$
$$ \mathbb{E}(\bar Z_n^m)={1\over n^m}\sum_{k=0}^n {n\choose k} k^m\left({1\over 2}\right)^n \to {1\over 2^m},$$
by the bounded convergence theorem.  
