High order central moments of a symmetric binomial variable Consider a random variable $X\sim B(n,\frac 12)$. I'm trying to estimate the asymptotic behaviour of its central moments $E((X-\frac n2)^r)$, where $r$ is even and in the range $\Omega(1)\leq r\leq O(n)$, and $n$ goes to $\infty$.
I've looked at inequalities for central moments of sums of independent variables, but they seem too general. I'd be very grateful if someone can point me in the right direction.
 A: I wonder if this perhaps-naive approach can help you. Let's shift by the mean, so consider $X = \sum_{j=1}^n X_j$ where each $X_j = 0.5$ or $-0.5$ independently with one-half probability each. So $\mathbb{E} X^r$ should be the value you're looking for. Now,
\begin{align}
  X^r &= \left(X_1 + \dots + X_n\right)^r  \\
  &= \sum_{\text{$j_1,\dots,j_r$}} ~~ \prod_{s=1}^r X_{j_s}
\end{align}
where the sum is over all vectors $(j_1,\dots,j_r)$ of indices in $\{1,\dots,n\}$.
Now the expectation of the sum is the sum of the expectations, so let's look at the expectation of one term:
\begin{align}
 \mathbb{E} \prod_{s=1}^r X_{j_s}
  &= \left(0.5^r\right) \mathbb{E} \prod_{s=1}^r \begin{cases} 1 & X_{j_s} > 0 \\ -1 & \text{otherwise} \end{cases} \\
  &= \left(0.5^r\right) \begin{cases} 1 & \text{each index appears in $\vec{j}$ an even number of times}  \\
           0 & \text{otherwise}. \end{cases}
\end{align}
(Why is this true? We split the product into a product over distinct indices $j$ of $X_j^{m(j)}$, where $m(j)$ is the multiplicity of $j$ in the multiset. The indices are independent Bernoullis, so the expectation of the product is the product of the expectations. Each distinct index's expected product is 1 if that index has even multiplicity and zero if odd.)
So if I've made no mistakes, the expectation you're looking for is equal to $0.5^r$ times the number of vectors in $\{1,\dots,n\}^r$ where each $j \in \{1,\dots,n\}$ appears in the vector an even number of times.
Example: For $r=2$, there are $n$ vectors of all-even multiplicity (one for each index, where that index appears twice), so we get $0.5^2*n = n/4$. For $r=4$, there are $n$ vectors containing just $1$ element and $6{n\choose 2}$ vectors containing two elements twice each (because there are six ways to arrange 2 indices into 4 slots, each taking two). So we get $0.5^4 \left(n + 6{n \choose 2}\right) = \frac{1}{16}\frac{2n + 6n^2 - 6n}{2} = \frac{3n^2}{16} - \frac{n}{8}$ (which is also correct -- it's nice to check).
I didn't get to think about how hard this is to estimate in general (I hope someone much more knowledgable in combinatorics can comment).
A: Let $M:=(E(X-n/2)^r)^{1/r}$. 
By Corollary 2 in [Latala], 
$M\sim S$, where $A\sim B$ means that $\frac1C\,B\le A\le C B$ for some universal positive constant $C$ and 
\begin{equation}
 S:=r\sup\{t(2n/r)^t\colon1/r\le t\le t_*\}, 
\end{equation}
where $t_*:=\frac12\wedge\frac nr$. It is not hard to see that $t(2n/r)^t$ increases in $t\in[1/r,t_*]$, and so, 
$S=rt_*(2n/r)^{t_*}$. 
Thus, 
$M\sim\sqrt{nr}$ if $r\le2n$ and $M\sim n(n/r)^{n/r}\sim n$ if $r\ge2n$. 
That is, $M\sim\sqrt{n(r\wedge n)}$. 
A: Symmetric Binomial $B(n,1/2)$ has sub-gaussian tails with variance-proxy $\sigma^2 = n/4$. Thus the growth of of the $r$-th moment is at most $(\mathbf{E}\|S\|^r)^{1/r} = O(\sqrt{r}\cdot \sigma) = O(\sqrt{r n})$.
The constant can be calculated by integrating the tails, see for example 
https://www.hse.ru/data/2016/11/24/1113029206/Concentration%20inequalities.pdf
