(This posting is a minor revision of my answer in Math.StackExchange

In this posting, we prove:

Theorem. Let $p > 3$ be a prime, and let $n = \frac{p-1}{2}$. Then

\[\sum_{k=1}^{\frac{p-1}{2}} \frac{1}{k} \binom{2k}{k} \equiv 0 \pmod{p}\]

For simplicity, we write $x = y + \mathcal{O}(p^r)$ if $ x \equiv y \pmod{p^r}$ holds.

Step 1. We begin by noting that, for each $k \in {0, 1, \ldots, p-1}$,

\[k! = \frac{(p-1)!}{\prod_{j=1}^{p-1-k} (p - j)} = \frac{(-1)^{p-k-1}(p-1)!}{(p-1-k)!} + \mathcal{O}(p)\]

holds. So, if $p$ is an odd prime and $n = \frac{p-1}{2}$, then

\[\begin{align*} \sum_{k=1}^{n} \frac{1}{k}\binom{2k}{k} &= 2 \sum_{k=1}^{n} \frac{(2k-1)!}{k!^2} \\ &= (-2) \sum_{k=1}^{n} \frac{(p-1)!}{k!^2(p-2k)!} + \mathcal{O}(p). \end{align*}\]

In light of this, the theorem immediately follows once we prove:

Claim. Let $p > 3$ be a prime. Then

\[\sum_{k \geq 0} \binom{p}{k,k,p-2k} = 1 + \mathcal{O}(p^2). \tag{1}\]

Here, we use the convention that $\frac{1}{i!} = 0$ if $i < 0$.

Step 2 - Proof of Claim. Let $S_p$ denote the LHS of $\text{(1)}$. Then we find that

\[\begin{align*} S_p = [x^p] (1 + x + x^2)^p. \end{align*}\]

Now we expand the RHS as a formal power series:

\[\begin{align*} (1 + x + x^2)^p &= \frac{(1-x^3)^p}{(1-x)^p} \\ &= \left( \sum_{j=0}^{p} (-1)^j \binom{p}{j} x^{3j} \right)\left( \sum_{k=0}^{\infty} \binom{p-1+k}{k} x^k \right) \\ &= \sum_{j=0}^{p} \sum_{k=0}^{\infty} (-1)^j \binom{p}{j} \binom{p-1+k}{k} x^{3j+k} \end{align*}\]

This shows that $S_p$ can be expanded as:

\[S_p = \sum_{\substack{j, k \geq 0 \\ 3j+k = p}} (-1)^j \binom{p}{j} \binom{p-1+k}{k} \tag{2}\]

In order to analyze the RHS of $\text{(2)}$, we observe that:

  • $\binom{p}{j} = \frac{p!}{j!(p-j)!} = \mathcal{O}(p)$ if and only if $0 < j < p$.
  • $\binom{p-1+k}{k} = \frac{(k+1) \cdots (k+p-1)}{(p-1)!} = \mathcal{O}(p)$ if and only if $p \nmid k$.
  • If $j \geq 1$ and $k = p-3j \geq 0$, then we must have $0 < k < p$. (This is the first place where the assumption $p > 3$ is used.) So by the previous observations, we have $(-1)^j \binom{p}{j} \binom{p-1+k}{k} = \mathcal{O}(p^2)$.

Consequently, we have

\[\begin{align*} S_p = \binom{2p-1}{p-1} + \mathcal{O}(p^2). \end{align*}\]

Then the desired claim follows by Wolstenholme’s theorem, completing the proof.