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$\newcommand\ep\varepsilon\newcommand\de\delta$Note that $$\sum_{k=0}^n k^a\binom nk=n^a2^n E\Big(\frac{X_n}n\Big)^a, \tag{1}\label{1}$$ where $X_n$ is a binomial random variable with parameters $n,1/2$. By the law of large numbers, $$\frac{X_n}n\to\frac12 \tag{2}\label{2}$$ in probability (as $n\to\infty$). So, by \eqref{1} and dominated convergence, for each real $a>0$, $$\sum_{k=0}^n k^a\binom nk\sim n^a2^n \Big(\frac12\Big)^a=n^a2^{n-a}. \tag{3}\label{3}$$

Details on \eqref{1}: We have $P(X_n=k)=\binom nk(\frac12)^n =2^{-n}\binom nk$. So, $$E\Big(\frac{X_n}n\Big)^a=\sum_{k=0}^n\Big(\frac kn\Big)^a P(X_n=k)=\sum_{k=0}^n\Big(\frac kn\Big)^a 2^{-n}\binom nk,$$ which yields \eqref{1}.

Details on \eqref{2}: By Chebyshev's inequality, for each real $\ep>0$, $$P\Big(\Big|\frac{X_n}n-\frac12\Big|\ge\ep\Big)\le\frac{1/(4n)}{\ep^2}\to0, \tag{4}\label{4} $$ which yields \eqref{2}.

Details on \eqref{3}: The dominated convergence theorem can be found in any standard textbook on probability (based on measure theory) and also in some books on mathematical statistics --see e.g. this, Exercise 5.2.22(ii); see also Theorem 5.2.5 there to infer from \eqref{2} that $(\frac{X_n}n)^a\to(\frac12)^a$ in probability.

In this case, it is easy to get \eqref{3} directly. Indeed, let $R_n:=\frac{X_n}n$ and $Y_n:=|R_n-\frac12|$. Then $0\le R_n\le1$ and hence for real $a>0$ $$\Big|R_n^a-\Big(\frac12\Big)^a\Big|\le a\Big|R_n-\frac12\Big|=aY_n.$$ So, $$E\Big|R_n^a-\Big(\frac12\Big)^a\Big|\le aEY_n. \tag{5}\label{5}$$ Next, $$EY_n=e_1+e_2,$$ where $$e_1:=EY_n\,1(Y_n\le\ep)\le\ep,$$ $$e_2:=EY_n\,1(Y_n>\ep)\le E\frac12\,1(Y_n>\ep) =\frac12\,P(Y_n>\ep)\to0,$$ by \eqref{4}. So, $EY_n\to0$ and hence, by \eqref{5}, $ER_n^a\to(\frac12)^a$. In view of \eqref{1}, this yields \eqref{3}.

To complete a previous discussion, let me present a probabilistic proof of the following more general

Fact: Let $f$ be any continuous real-valued function on $[0,1]$. Consider the Bernstein polynomial
$$f_n(p):=\sum_{k=0}^n f(k/n)p^k(1-p)^{n-k}.$$ Then $f_n\to f$ uniformly on $[0,1]$.

Proof: Note that $f_n(p)=Ef(Y_{n,p})$, where $Y_{n,p}:=X_{n,p}/n$ and $X_{n,p}$ is a binomial random variable with parameters $n,p$. Take any real $\ep>0$. Then, because $f$ is uniformly continuous on the compact set $[0,1]$, there is some real $\de>0$ such that for all $x,y$ in $[0,1]$ such that $|x-y|\le\de$ we have $|f(x)-f(y)|\le\ep$. So, $|f(Y_{n,p})-f(p)|\le\ep$ on the event $\{|Y_{n,p}-p|\le\de\}$. Therefore and because $|f|\le M$ for some real $M$, we have $$|f_n(p)-f(p)|\le E|f(Y_{n,p})-f(p)|\le e_1+e_2,$$ where $$e_1:=E|f(Y_{n,p})-f(p)|\,1(|Y_{n,p}-p|\le\de)\le\ep,$$ $$e_2:=E|f(Y_{n,p})-f(p)|\,1(|Y_{n,p}-p|>\de) \le2MP(|Y_{n,p}-p|>\de)\le2M\frac{E(Y_{n,p}-p)^2}{\de^2} \le2M\frac{p(1-p)}{n\de^2} \le\frac M{2n\de^2}.$$ So, $$\|f_n-f\|=\max_{p\in[0,1]}|f_n(p)-f(p)|\le\ep+\frac M{2n\de^2}.$$ Thus, $$\limsup_n\|f_n-f\|\le\ep,$$ for each real $\ep>0$. $\quad\Box$

The OP is about the special case of the just proved fact, when $f(p)=p^\alpha$ (with positive $\alpha$) and the fixed $p=1/2$.

$\newcommand\ep\varepsilon\newcommand\de\delta$Note that $$\sum_{k=0}^n k^a\binom nk=n^a2^n E\Big(\frac{X_n}n\Big)^a, \tag{1}\label{1}$$ where $X_n$ is a binomial random variable with parameters $n,1/2$. By the law of large numbers, $$\frac{X_n}n\to\frac12 \tag{2}\label{2}$$ in probability (as $n\to\infty$). So, by \eqref{1} and dominated convergence, for each real $a>0$, $$\sum_{k=0}^n k^a\binom nk\sim n^a2^n \Big(\frac12\Big)^a=n^a2^{n-a}. \tag{3}\label{3}$$

Details on \eqref{1}: We have $P(X_n=k)=\binom nk(\frac12)^n =2^{-n}\binom nk$. So, $$E\Big(\frac{X_n}n\Big)^a=\sum_{k=0}^n\Big(\frac kn\Big)^a P(X_n=k)=\sum_{k=0}^n\Big(\frac kn\Big)^a 2^{-n}\binom nk,$$ which yields \eqref{1}.

Details on \eqref{2}: By Chebyshev's inequality, for each real $\ep>0$, $$P\Big(\Big|\frac{X_n}n-\frac12\Big|\ge\ep\Big)\le\frac{1/(4n)}{\ep^2}\to0, \tag{4}\label{4} $$ which yields \eqref{2}.

Details on \eqref{3}: The dominated convergence theorem can be found in any standard textbook on probability (based on measure theory) and also in some books on mathematical statistics --see e.g. this, Exercise 5.2.22(ii); see also Theorem 5.2.5 there to infer from \eqref{2} that $(\frac{X_n}n)^a\to(\frac12)^a$ in probability.

In this case, it is easy to get \eqref{3} directly. Indeed, let $R_n:=\frac{X_n}n$ and $Y_n:=|R_n-\frac12|$. Then $0\le R_n\le1$ and hence for real $a>0$ $$\Big|R_n^a-\Big(\frac12\Big)^a\Big|\le a\Big|R_n-\frac12\Big|=aY_n.$$ So, $$E\Big|R_n^a-\Big(\frac12\Big)^a\Big|\le aEY_n. \tag{5}\label{5}$$ Next, $$EY_n=e_1+e_2,$$ where $$e_1:=EY_n\,1(Y_n\le\ep)\le\ep,$$ $$e_2:=EY_n\,1(Y_n>\ep)\le E\frac12\,1(Y_n>\ep) =\frac12\,P(Y_n>\ep)\to0,$$ by \eqref{4}. So, $EY_n\to0$ and hence, by \eqref{5}, $ER_n^a\to(\frac12)^a$. In view of \eqref{1}, this yields \eqref{3}.

To complete a previous discussion, let me present a probabilistic proof of the following more general

Fact: Let $f$ be any continuous real-valued function on $[0,1]$. Consider the Bernstein polynomial
$$f_n(p):=\sum_{k=0}^n f(k/n)\binom nk p^k(1-p)^{n-k}.$$ Then $f_n\to f$ uniformly on $[0,1]$.

Proof: Note that $f_n(p)=Ef(Y_{n,p})$, where $Y_{n,p}:=X_{n,p}/n$ and $X_{n,p}$ is a binomial random variable with parameters $n,p$. Take any real $\ep>0$. Then, because $f$ is uniformly continuous on the compact set $[0,1]$, there is some real $\de>0$ such that for all $x,y$ in $[0,1]$ such that $|x-y|\le\de$ we have $|f(x)-f(y)|\le\ep$. So, $|f(Y_{n,p})-f(p)|\le\ep$ on the event $\{|Y_{n,p}-p|\le\de\}$. Therefore and because $|f|\le M$ for some real $M$, we have $$|f_n(p)-f(p)|\le E|f(Y_{n,p})-f(p)|\le e_1+e_2,$$ where $$e_1:=E|f(Y_{n,p})-f(p)|\,1(|Y_{n,p}-p|\le\de)\le\ep,$$ $$e_2:=E|f(Y_{n,p})-f(p)|\,1(|Y_{n,p}-p|>\de) \le2MP(|Y_{n,p}-p|>\de)\le2M\frac{E(Y_{n,p}-p)^2}{\de^2} \le2M\frac{p(1-p)}{n\de^2} \le\frac M{2n\de^2}.$$ So, $$\|f_n-f\|=\max_{p\in[0,1]}|f_n(p)-f(p)|\le\ep+\frac M{2n\de^2}.$$ Thus, $$\limsup_n\|f_n-f\|\le\ep,$$ for each real $\ep>0$. $\quad\Box$

The OP is about the special case of the just proved fact, when $f(p)=p^\alpha$ (with positive $\alpha$) and the fixed $p=1/2$.

$\newcommand\ep\varepsilon\newcommand\de\delta$Note that $$\sum_{k=0}^n k^a\binom nk=n^a2^n E\Big(\frac{X_n}n\Big)^a, \tag{1}\label{1}$$ where $X_n$ is a binomial random variable with parameters $n,1/2$. By the law of large numbers, $$\frac{X_n}n\to\frac12 \tag{2}\label{2}$$ in probability (as $n\to\infty$). So, by \eqref{1} and dominated convergence, for each real $a>0$, $$\sum_{k=0}^n k^a\binom nk\sim n^a2^n \Big(\frac12\Big)^a=n^a2^{n-a}. \tag{3}\label{3}$$

Details on \eqref{1}: We have $P(X_n=k)=\binom nk(\frac12)^n =2^{-n}\binom nk$. So, $$E\Big(\frac{X_n}n\Big)^a=\sum_{k=0}^n\Big(\frac kn\Big)^a P(X_n=k)=\sum_{k=0}^n\Big(\frac kn\Big)^a 2^{-n}\binom nk,$$ which yields \eqref{1}.

Details on \eqref{2}: By Chebyshev's inequality, for each real $\ep>0$, $$P\Big(\Big|\frac{X_n}n-\frac12\Big|\ge\ep\Big)\le\frac{1/(4n)}{\ep^2}\to0, \tag{4}\label{4} $$ which yields \eqref{2}.

Details on \eqref{3}: The dominated convergence theorem can be found in any standard textbook on probability (based on measure theory) and also in some books on mathematical statistics --see e.g. this, Exercise 5.2.22(ii); see also Theorem 5.2.5 there to infer from \eqref{2} that $(\frac{X_n}n)^a\to(\frac12)^a$ in probability.

In this case, it is easy to get \eqref{3} directly. Indeed, let $R_n:=\frac{X_n}n$ and $Y_n:=|R_n-\frac12|$. Then $0\le R_n\le1$ and hence for real $a>0$ $$\Big|R_n^a-\Big(\frac12\Big)^a\Big|\le a\Big|R_n-\frac12\Big|=aY_n.$$ So, $$E\Big|R_n^a-\Big(\frac12\Big)^a\Big|\le aEY_n. \tag{5}\label{5}$$ Next, $$EY_n=e_1+e_2,$$ where $$e_1:=EY_n\,1(Y_n\le\ep)\le\ep,$$ $$e_2:=EY_n\,1(Y_n>\ep)\le E\frac12\,1(Y_n>\ep) =\frac12\,P(Y_n>\ep)\to0,$$ by \eqref{4}. So, $EY_n\to0$ and hence, by \eqref{5}, $ER_n^a\to(\frac12)^a$. In view of \eqref{1}, this yields \eqref{3}.

To complete a previous discussion, let me present a probabilistic proof of the following more general

Fact: Let $f$ be any continuous real-valued function on $[0,1]$. Consider the Bernstein polynomial
$$f_n(p):=\sum_{k=0}^n f(k/n)p^k(1-p)^{n-k}.$$ Then $f_n\to f$ uniformly on $[0,1]$.

Proof: Note that $f_n(p)=Ef(X_{n,p})$, where $X_{n,p}$ is a binomial random variable with parameters $n,p$. Take any real $\ep>0$. Then, because $f$ is uniformly continuous on the compact set $[0,1]$, there is some real $\de>0$ such that for all $x,y$ in $[0,1]$ such that $|x-y|\le\de$ we have $|f(x)-f(y)|\le\ep$. So, $|f(X_{n,p})-f(p)|\le\ep$ on the event $\{|X_{n,p}-p|\le\de\}$. Therefore and because $|f|\le M$ for some real $M$, we have $$|f_n(p)-f(p)|\le E|f(X_{n,p})-f(p)|\le e_1+e_2,$$ where $$e_1:=E|f(X_{n,p})-f(p)|\,1(|X_{n,p}-p|\le\de)\le\ep,$$ $$e_2:=E|f(X_{n,p})-f(p)|\,1(|X_{n,p}-p|>\de) \le2MP(|X_{n,p}-p|>\de)\le2M\frac{p(1-p)}{\de^2} \le\frac M{2n\de^2}.$$ So, $$\|f_n-f\|=\max_{p\in[0,1]}|f_n(p)-f(p)|\le\ep+\frac M{2n\de^2}.$$ Thus, $$\limsup_n\|f_n-f\|\le\ep,$$ for each real $\ep>0$. $\quad\Box$

The OP is about the special case of the just proved fact, when $f(p)=p^\alpha$ (with positive $\alpha$) and the fixed $p=1/2$.

$\newcommand\ep\varepsilon\newcommand\de\delta$Note that $$\sum_{k=0}^n k^a\binom nk=n^a2^n E\Big(\frac{X_n}n\Big)^a, \tag{1}\label{1}$$ where $X_n$ is a binomial random variable with parameters $n,1/2$. By the law of large numbers, $$\frac{X_n}n\to\frac12 \tag{2}\label{2}$$ in probability (as $n\to\infty$). So, by \eqref{1} and dominated convergence, for each real $a>0$, $$\sum_{k=0}^n k^a\binom nk\sim n^a2^n \Big(\frac12\Big)^a=n^a2^{n-a}. \tag{3}\label{3}$$

Details on \eqref{1}: We have $P(X_n=k)=\binom nk(\frac12)^n =2^{-n}\binom nk$. So, $$E\Big(\frac{X_n}n\Big)^a=\sum_{k=0}^n\Big(\frac kn\Big)^a P(X_n=k)=\sum_{k=0}^n\Big(\frac kn\Big)^a 2^{-n}\binom nk,$$ which yields \eqref{1}.

Details on \eqref{2}: By Chebyshev's inequality, for each real $\ep>0$, $$P\Big(\Big|\frac{X_n}n-\frac12\Big|\ge\ep\Big)\le\frac{1/(4n)}{\ep^2}\to0, \tag{4}\label{4} $$ which yields \eqref{2}.

Details on \eqref{3}: The dominated convergence theorem can be found in any standard textbook on probability (based on measure theory) and also in some books on mathematical statistics --see e.g. this, Exercise 5.2.22(ii); see also Theorem 5.2.5 there to infer from \eqref{2} that $(\frac{X_n}n)^a\to(\frac12)^a$ in probability.

In this case, it is easy to get \eqref{3} directly. Indeed, let $R_n:=\frac{X_n}n$ and $Y_n:=|R_n-\frac12|$. Then $0\le R_n\le1$ and hence for real $a>0$ $$\Big|R_n^a-\Big(\frac12\Big)^a\Big|\le a\Big|R_n-\frac12\Big|=aY_n.$$ So, $$E\Big|R_n^a-\Big(\frac12\Big)^a\Big|\le aEY_n. \tag{5}\label{5}$$ Next, $$EY_n=e_1+e_2,$$ where $$e_1:=EY_n\,1(Y_n\le\ep)\le\ep,$$ $$e_2:=EY_n\,1(Y_n>\ep)\le E\frac12\,1(Y_n>\ep) =\frac12\,P(Y_n>\ep)\to0,$$ by \eqref{4}. So, $EY_n\to0$ and hence, by \eqref{5}, $ER_n^a\to(\frac12)^a$. In view of \eqref{1}, this yields \eqref{3}.

To complete a previous discussion, let me present a probabilistic proof of the following more general

Fact: Let $f$ be any continuous real-valued function on $[0,1]$. Consider the Bernstein polynomial
$$f_n(p):=\sum_{k=0}^n f(k/n)p^k(1-p)^{n-k}.$$ Then $f_n\to f$ uniformly on $[0,1]$.

Proof: Note that $f_n(p)=Ef(Y_{n,p})$, where $Y_{n,p}:=X_{n,p}/n$ and $X_{n,p}$ is a binomial random variable with parameters $n,p$. Take any real $\ep>0$. Then, because $f$ is uniformly continuous on the compact set $[0,1]$, there is some real $\de>0$ such that for all $x,y$ in $[0,1]$ such that $|x-y|\le\de$ we have $|f(x)-f(y)|\le\ep$. So, $|f(Y_{n,p})-f(p)|\le\ep$ on the event $\{|Y_{n,p}-p|\le\de\}$. Therefore and because $|f|\le M$ for some real $M$, we have $$|f_n(p)-f(p)|\le E|f(Y_{n,p})-f(p)|\le e_1+e_2,$$ where $$e_1:=E|f(Y_{n,p})-f(p)|\,1(|Y_{n,p}-p|\le\de)\le\ep,$$ $$e_2:=E|f(Y_{n,p})-f(p)|\,1(|Y_{n,p}-p|>\de) \le2MP(|Y_{n,p}-p|>\de)\le2M\frac{E(Y_{n,p}-p)^2}{\de^2} \le2M\frac{p(1-p)}{n\de^2} \le\frac M{2n\de^2}.$$ So, $$\|f_n-f\|=\max_{p\in[0,1]}|f_n(p)-f(p)|\le\ep+\frac M{2n\de^2}.$$ Thus, $$\limsup_n\|f_n-f\|\le\ep,$$ for each real $\ep>0$. $\quad\Box$

The OP is about the special case of the just proved fact, when $f(p)=p^\alpha$ (with positive $\alpha$) and the fixed $p=1/2$.

$\newcommand\ep\varepsilon\newcommand\de\delta$Note that $$\sum_{k=0}^n k^a\binom nk=n^a2^n E\Big(\frac{X_n}n\Big)^a, \tag{1}\label{1}$$ where $X_n$ is a binomial random variable with parameters $n,1/2$. By the law of large numbers, $$\frac{X_n}n\to\frac12 \tag{2}\label{2}$$ in probability (as $n\to\infty$). So, by \eqref{1} and dominated convergence, for each real $a>0$, $$\sum_{k=0}^n k^a\binom nk\sim n^a2^n \Big(\frac12\Big)^a=n^a2^{n-a}. \tag{3}\label{3}$$

Details on \eqref{1}: We have $P(X_n=k)=\binom nk(\frac12)^n =2^{-n}\binom nk$. So, $$E\Big(\frac{X_n}n\Big)^a=\sum_{k=0}^n\Big(\frac kn\Big)^a P(X_n=k)=\sum_{k=0}^n\Big(\frac kn\Big)^a 2^{-n}\binom nk,$$ which yields \eqref{1}.

Details on \eqref{2}: By Chebyshev's inequality, for each real $\ep>0$, $$P\Big(\Big|\frac{X_n}n-\frac12\Big|\ge\ep\Big)\le\frac{1/(4n)}{\ep^2}\to0, \tag{4}\label{4} $$ which yields \eqref{2}.

Details on \eqref{3}: The dominated convergence theorem can be found in any standard textbook on probability (based on measure theory) and also in some books on mathematical statistics --see e.g. this, Exercise 5.2.22(ii); see also Theorem 5.2.5 there to infer from \eqref{2} that $(\frac{X_n}n)^a\to(\frac12)^a$ in probability.

In this case, it is easy to get \eqref{3} directly. Indeed, let $R_n:=\frac{X_n}n$ and $Y_n:=|R_n-\frac12|$. Then $0\le R_n\le1$ and hence for real $a>0$ $$\Big|R_n^a-\Big(\frac12\Big)^a\Big|\le a\Big|R_n-\frac12\Big|=aY_n.$$ So, $$E\Big|R_n^a-\Big(\frac12\Big)^a\Big|\le aEY_n. \tag{5}\label{5}$$ Next, $$EY_n=e_1+e_2,$$ where $$e_1:=EY_n\,1(Y_n\le\ep)\le\ep,$$ $$e_2:=EY_n\,1(Y_n>\ep)\le E\frac12\,1(Y_n>\ep) =\frac12\,P(Y_n>\ep)\to0,$$ by \eqref{4}. So, $EY_n\to0$ and hence, by \eqref{5}, $ER_n^a\to(\frac12)^a$. In view of \eqref{1}, this yields \eqref{3}.

To complete a previous discussion, let me present a probabilistic proof of the following more general

Fact: Let $f$ be any continuous real-valued function on $[0,1]$. Consider the Bernstein polynomial
$$f_n(p):=\sum_{k=0}^n f(k/n)p^k(1-p)^{n-k}.$$ Then $f_n\to f$ uniformly on $[0,1]$.

Proof: Note that $f_n(p)=Ef(X_{n,p})$, where $X_{n,p}$ is a binomial random variable with parameters $n,p$. Take any real $\ep>0$. Then, because $f$ is uniformly continuous on the compact set $[0,1]$, there is some real $\de>0$ such that for all $x,y$ in $[0,1]$ such that $|x-y|\le\de$ we have $|f(x)-f(y)|\le\ep$. So, $|f(X_{n,p})-f(p)|\le\ep$ on the event $\{|X_{n,p}-p|\le\de\}$. Therefore and because $|f|\le M$ for some real $M$, we have $$|f_n(p)-f(p)|\le E|f(X_{n,p})-f(p)|\le e_1+e_2,$$ where $$e_1:=E|f(X_{n,p})-f(p)|\,1(|X_{n,p}-p|\le\de)\le\ep,$$ $$e_2:=E|f(X_{n,p})-f(p)|\,1(|X_{n,p}-p|>\de) \le2MP(|X_{n,p}-p|>\de)\le2M\frac{p(1-p)}{\de^2} \le\frac M{2n\de^2}.$$ So, $$\|f_n-f\|=\max_{p\in[0,1]}|f_n(p)-f(p)|\le\ep+\frac M{2n\de^2}.$$ Thus, $$\limsup_n\|f_n-f\|\le\ep,$$ for each real $\ep>0$. $\quad\Box$

$\newcommand\ep\varepsilon\newcommand\de\delta$Note that $$\sum_{k=0}^n k^a\binom nk=n^a2^n E\Big(\frac{X_n}n\Big)^a, \tag{1}\label{1}$$ where $X_n$ is a binomial random variable with parameters $n,1/2$. By the law of large numbers, $$\frac{X_n}n\to\frac12 \tag{2}\label{2}$$ in probability (as $n\to\infty$). So, by \eqref{1} and dominated convergence, for each real $a>0$, $$\sum_{k=0}^n k^a\binom nk\sim n^a2^n \Big(\frac12\Big)^a=n^a2^{n-a}. \tag{3}\label{3}$$

Details on \eqref{1}: We have $P(X_n=k)=\binom nk(\frac12)^n =2^{-n}\binom nk$. So, $$E\Big(\frac{X_n}n\Big)^a=\sum_{k=0}^n\Big(\frac kn\Big)^a P(X_n=k)=\sum_{k=0}^n\Big(\frac kn\Big)^a 2^{-n}\binom nk,$$ which yields \eqref{1}.

Details on \eqref{2}: By Chebyshev's inequality, for each real $\ep>0$, $$P\Big(\Big|\frac{X_n}n-\frac12\Big|\ge\ep\Big)\le\frac{1/(4n)}{\ep^2}\to0, \tag{4}\label{4} $$ which yields \eqref{2}.

Details on \eqref{3}: The dominated convergence theorem can be found in any standard textbook on probability (based on measure theory) and also in some books on mathematical statistics --see e.g. this, Exercise 5.2.22(ii); see also Theorem 5.2.5 there to infer from \eqref{2} that $(\frac{X_n}n)^a\to(\frac12)^a$ in probability.

In this case, it is easy to get \eqref{3} directly. Indeed, let $R_n:=\frac{X_n}n$ and $Y_n:=|R_n-\frac12|$. Then $0\le R_n\le1$ and hence for real $a>0$ $$\Big|R_n^a-\Big(\frac12\Big)^a\Big|\le a\Big|R_n-\frac12\Big|=aY_n.$$ So, $$E\Big|R_n^a-\Big(\frac12\Big)^a\Big|\le aEY_n. \tag{5}\label{5}$$ Next, $$EY_n=e_1+e_2,$$ where $$e_1:=EY_n\,1(Y_n\le\ep)\le\ep,$$ $$e_2:=EY_n\,1(Y_n>\ep)\le E\frac12\,1(Y_n>\ep) =\frac12\,P(Y_n>\ep)\to0,$$ by \eqref{4}. So, $EY_n\to0$ and hence, by \eqref{5}, $ER_n^a\to(\frac12)^a$. In view of \eqref{1}, this yields \eqref{3}.

To complete a previous discussion, let me present a probabilistic proof of the following more general

Fact: Let $f$ be any continuous real-valued function on $[0,1]$. Consider the Bernstein polynomial
$$f_n(p):=\sum_{k=0}^n f(k/n)p^k(1-p)^{n-k}.$$ Then $f_n\to f$ uniformly on $[0,1]$.

Proof: Note that $f_n(p)=Ef(X_{n,p})$, where $X_{n,p}$ is a binomial random variable with parameters $n,p$. Take any real $\ep>0$. Then, because $f$ is uniformly continuous on the compact set $[0,1]$, there is some real $\de>0$ such that for all $x,y$ in $[0,1]$ such that $|x-y|\le\de$ we have $|f(x)-f(y)|\le\ep$. So, $|f(X_{n,p})-f(p)|\le\ep$ on the event $\{|X_{n,p}-p|\le\de\}$. Therefore and because $|f|\le M$ for some real $M$, we have $$|f_n(p)-f(p)|\le E|f(X_{n,p})-f(p)|\le e_1+e_2,$$ where $$e_1:=E|f(X_{n,p})-f(p)|\,1(|X_{n,p}-p|\le\de)\le\ep,$$ $$e_2:=E|f(X_{n,p})-f(p)|\,1(|X_{n,p}-p|>\de) \le2MP(|X_{n,p}-p|>\de)\le2M\frac{p(1-p)}{\de^2} \le\frac M{2n\de^2}.$$ So, $$\|f_n-f\|=\max_{p\in[0,1]}|f_n(p)-f(p)|\le\ep+\frac M{2n\de^2}.$$ Thus, $$\limsup_n\|f_n-f\|\le\ep,$$ for each real $\ep>0$. $\quad\Box$

The OP is about the special case of the just proved fact, when $f(p)=p^\alpha$ (with positive $\alpha$) and the fixed $p=1/2$.