An asymptotic expression of sum of powers of binomial 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. The proof provided in the book is too simple. It says details can be found in Jordan's Cours d'Analyse, Vol.2, 3rd Ed, pp. 218-221. However, I cannot find this edition online, and what's worse, there is not any English translations. Can anyone give a proof in detail?
 A: 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:
$$\mathbb{E}[X]=\frac{n}{4}, \qquad \operatorname{Var}[X]=\frac{n}{4}$$
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:
$$\int_{-\infty}^{+\infty}\exp\left(-\frac{2kx^2}{n}\right)\,dx = \sqrt{\frac{\pi n}{2k}}.\tag{3}$$
A: This is not an answer, but is too long for the usual comment format.
I have access to the reference given in the OP, so I looked it up,
but I was disappointed.
In those pages 218 to 221, C. Jordan nowhere considers sums 
$\sum_{\nu} \binom{n}{\nu}^k$ for $k$ other than $1$. The strongest result 
shown there is a form of the CLT : Jordan shows that for $p\in[0,1]$, 
if $(\lambda_{n})$ is an integer sequence such that 
$\frac{\lambda_n}{n^a}$ is bounded for some $a<1$, then 
$$
\sum_{\nu=\frac{n}{2}-\lambda_n}^{\frac{n}{2}+\lambda_n}
\binom{n}{\nu}p^\nu(1-p)^{n-\nu} \sim_{n\to+\infty} 
\frac{1}{\sqrt{2\pi p(1-p)n}} \int_{-\lambda_n}^{\lambda_n} e^{-\frac{x^2}{2p(1-p)n}}dx
$$
The key ingredient of the proof is that as $\lambda_n$ is sufficiently small
compared to $n$, we can use a Stirling approximation $\log(t!)\sim (t+\frac{1}{2})\log(t)-t+\frac{\log(2\pi)}{2}$ for all the integers $t$ we need.
The only way this can help in my view is that it suggests showing something
like
$$
\sum_{\nu =0}^n \binom{n}{\nu}^k \sim \sum_{\nu =\frac{n}{2}-n^a}^{\frac{n}{2}+n^a}\binom{n}{\nu}^k 
$$
for some $a<1$ depending perhaps on $k$. That’s still a big "if".
