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Last edited by hbghlyj at 2023-5-30 15:14:00math.stackexchange.com/questions/976777/an-as … inomial-coefficients
Let $k$ be a fixed positive number and $n$ an integer increasing to infinity. Then $$\sum_{\nu =0}^n \binom{n}{\nu}^k \sim \frac{2^{kn}}{\sqrt{k}} \left( \frac{2}{\pi n} \right)^{\frac{k-1}{2}}.$$This is from Polya's Problems and Theorems in Analysis, Vol. 1, Part II, Problem 40.
That kind of asymptotics follows from the Central Limit Theorem. If we consider the binomial random variable $X=B(n,1/2)$ as the sum of $n$ independent Bernoulli trials, we have:
$$\mathbb{E}[X]=\frac{n}{2}, \qquad \operatorname{Var}[X]=\frac{n}{4}$$
from which the approximation:
$$\frac{1}{2^n}\binom{n}{n/2+r}\approx \sqrt{\frac{2}{n\pi}}\exp\left(-\frac{2r^2}{n}\right).\tag{1}$$
By considering the $k$-th power of both terms and summing over $r\in[-n/2,n/2]$ (the main contribute is clearly given by the central binomial coefficient and its neighbours) we get:
$$\sum_{r=-n/2}^{n/2}\binom{n}{n/2+r}^k \approx 2^{kn}\left(\frac{2}{\pi n}\right)^{\frac{k}{2}}\sum_{r=-n/2}^{n/2}\exp\left(-\frac{2kr^2}{n}\right)\tag{2}$$
and the claim follows from approximating the last sum with:
$$\int_{-\infty}^{+\infty}\exp\left(-\frac{2kx^2}{n}\right)\,dx = \sqrt{\frac{\pi n}{2k}}.\tag{3}$$
See also: kuing.cjhb.site/forum.php?mod=viewthread&tid=9266 |
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