A binomial sum is divisible by p^2 This is a question I have since longer time, but I have absolutely no idea how to proceed on it.
Let $p>3$ be a prime. Prove that $\displaystyle\sum\limits_{k=1}^{p-1}\frac{1}{k}\binom{2k}{k}\equiv 0\mod p^2$. Here, we work in $\mathbb{Z}_{\mathbb{Z}\setminus p\mathbb{Z}}$ (that is, $\mathbb{Z}$ localized at all numbers not divisible by $p$).
I know that it is $0\mod p$ (though I can't find the reference at the moment; it was some hard olympiad problem on MathLinks). The $0\mod p^2$ assertion is backed up by computation for all $p<100$. I am sorry if this is trivial or known. I would be delighted to see a combinatorial proof (= finding a binomial identity which reduces to the above when computed $\mod p^2$). Some number-theoretical arguments would be nice, too. However, I fear that if you use analytic number theory, I will not understand a single word.
EDIT: Epic fail at question title fixed.
 A: For the asymptotics: begin with the well-known generating function
$$ \sum_{k \ge 0} {2k \choose k} z^k = (1-4z)^{-1/2}. $$
We can remove the first term to get
$$ \sum_{k \ge 1} {2k \choose k} z^k = (1-4z)^{-1/2} - 1. $$
Dividing by $z$, and then integrating with respect to $z$, turns $z^k$ into $z^k/k$.
$$ \sum_{k \ge 1} {2k \choose k} {1 \over k} z^k = -2 \log (1 + \sqrt{1-4z}) + 2 \log 2. $$
(The $2 \log 2$ is a constant of integration.)  
Finally, let $f(n) = \sum_{k=1}^n {1 \over k} {2k \choose k}$.  These are the partial sums of the previous sequence, so we have
$$ F(z) = \sum_{n \ge 1}  f(n) z^n = {-2 \over 1-z} \log( 1 + \sqrt{1-4z} ) $$
We can find the asymptotics of the coefficients of this generating function by singularity analysis.  The singularity closest to the origin is at $z = 1/4$.  Near $z = 1/4$,
$$ F(z) \sim {-8 \over 3} \sqrt{1-4z} $$
and by a transfer theorem (probably originally due to Flajolet and Odlyzko and in the Flajolet-Sedgewick book Analytic Combinatorics, although it's too late to look up the details right now) we get
$$ f(n) = [z^n] F(z) \sim {-8 \over 3} [z^n] \sqrt{1-4z} $$
Finally, $[z^n] \sqrt{1-4z} = -{2 \over n} {2n-2 \choose n-1}$, and so by Stirling's formula, as $n \to \infty$
$$ f(n) \sim {-8 \over 3} {-2 \over n} {4^{n-1} \over \sqrt{\pi n}} = {4 \over 3} {4^n \over \sqrt{\pi n^3}}. $$
Finally, because I was somewhat awkward in my indexing, we have to take $p = n-1$, and so
$$ f(p) \sim {1 \over 3} {4^p \over \sqrt{\pi p^3}} $$.
As has been observed by John Mangual and Mariano Suárez-Alvarez, this is about one-third of the $p$th Catalan number.  Examining the singularity near $z = 1/4$ more closely would lead to higher-order terms.
A: Your binomial sum is divisible by $p^2$ as is shown in a recent paper of Sun and Tauraso:
https://arxiv.org/abs/0805.0563
Even more remarkably, they compute that it's equivalent to $\frac{8}{9}p^2B_{p-3} \bmod p^3$, as well as various other congruences for the corresponding alternating sum.
