# Evaluating 'combinatorial' sum

Help me please to calculate the following sum. I have seen such kind of formulas in the papers related to combinatorics, specifically 'trees'. I am curious how to calculate or approximate this sum: Let $n \in N$, $q\geq 2$ $$\sum_{m=-n}^n m^q {n \choose (m+n)/2}=\Gamma(n+1)\sum_{m=-n}^nm^q\frac{1}{\Gamma(n/2-m/2+1)\Gamma(n/2+m/2+1)}$$

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This sum involves terms like $n\choose{1/2}$, which needs some defining. – Gerry Myerson May 1 '12 at 2:51
$\sum\limits_{m=0}^n (2m-n)^q \binom{n}{m}$ might be a better representation... – J. M. May 1 '12 at 2:55
@Gerry Myerson: sorry, I am not sure what kind of definition I should do here... I ' ve got this formula after few substitutions... Anyway, using Stirling's approximation formula, one can get $\sum_{m=-n}^n \frac{2^{n+1}m^qn^{n+1/2}}{(m+n)^{(m+n+1)/2}(n-m)^{(n-m+1)/2}}$. but, it did not help... – Nick G.H. May 1 '12 at 3:09
@J.M. : I am not sure how it helps... I've never seen such formulas before. could you please provide any book or paper... I would appresiate it. Thank you. – Nick G.H. May 1 '12 at 3:12
Well, you can define $n \choose (m+n)/2$ using the Gamma function. Note that for nonnegative integers $k$, $\Gamma(k+1/2) = (2k+1)!! \sqrt{\pi}/2^k$ – Robert Israel May 1 '12 at 5:42
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## 1 Answer

It seems the idea is to consider the sum over $m=-n+2k$ for $0\leqslant k\leqslant n$, which guarantees that every $k=\frac12(m+n)$ is an integer. Thus, the sum with parameter $q$ is $$\color{red}{s_n(q)=2^n\sum_{k=0}^n\binom{n}{k}2^{-n}(2k-n)^q}.$$ If $q$ is odd, $s_n(q)=0$ thanks to the symmetry $k\to n-k$. From now on, assume that $q$ is even. Note that $$s_n(q)=2^n\mathrm E((2X_n-n)^q),$$ where $X_n$ is the sum of $n$ i.i.d. Bernoulli random variables with parameter $\frac12$. Define some random variables $G_n$ by the identity $X_n=\frac12n+\frac12\sqrt{n}G_n$, then $$s_n(q)=2^nn^{q/2}\mathrm E((G_n)^q).$$ The central limit theorem asserts that $G_n$ converges in distribution to a standard normal random variable $G$. For binomial random variables, it happens that the moments of $G_n$ also converge to the moments of $G$, which are well known. Finally, for every even nonnegative integer $q$, $$\color{red}{\lim\limits_{n\to\infty}2^{-n}n^{-q/2}s_n(q)=\mathrm E(G^q)},$$ with $\mathrm E(G^q)=0$ if $q$ is odd and $\mathrm E(G^q)=(q-1)!!$ if $q$ is even.

Edit: Likewise, one can consider, for every nonnegative real number $q$, $$\color{green}{t_n(q)=\sum_{k=0}^n\binom{n}{k}\cdot|2k-n|^q}.$$ The same reasoning yields $$\color{green}{\lim\limits_{n\to\infty}2^{-n}n^{-q/2}t_n(q)=\mathrm E(|G|^q)=\frac{2^{q/2}}{\sqrt{\pi}}\Gamma\left(\frac{q+1}2\right)}.$$

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 Thank you very much, Didier, For your answer. I've never seen such kind of methods before. I would like to study it deeply. Could you provide any literature which would give me the similar examples? Thank you very much. – Nick G.H. May 6 '12 at 3:21 A famous example is the constructive proof of the Stoneâ€“Weierstrass approximation theorem using Bernstein polynomials and the (weak) law of large numbers. – Did May 6 '12 at 8:02 @did: its a very interesting solution. But I do not get how did you do the first 'changeover'? For which $n$, $k$? Thank you. – Michael Jun 5 '12 at 12:49 @Michael: Sorry but I have no idea what you are talking about. Could you make your question more precise? – Did Jun 5 '12 at 15:36 @Michael: Additionally, simply for correction you should have mentioned this. – Did Jun 5 '12 at 15:38
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