Let $p \equiv 1 \pmod 4$ be a prime and $E_n$ denote the $n$-th Euler number. While investigating $E_{p-1} \pmod{p^2}$ I have encountered this summation (modulo $p$) \begin{align*} \sum_{k =1}^{\frac{p-3}{2}} \frac{a_k}{2k+1} \pmod p. \end{align*} where $a_k = \frac{1\cdot 3\cdots (2k-1)}{2\cdot 4 \cdots 2k}$.

Another way to express this sum (this is how I first saw it) is \begin{align} \sum_{k = 1}^{\frac{p-3}{2}} {2k \choose k} \frac{1}{4^k(2k + 1)}. \end{align}

Like the questions says, I am wondering if a simple expression is known for this sum, or if sums like it have been studied before. Any references or insights would be appreciated!

Some observations: Using Wilson's theorem, we have \begin{align} \frac{1\cdot 3\cdots (p-2)}{2\cdot 4 \cdots (p-1)} \equiv \frac{-1}{(2\cdot 4 \cdots (p-1))^2} \equiv \frac{-1}{2^{p-1}\left(\frac{p-1}{2}\right)^2!} \equiv 1 \pmod p \end{align} ($p \equiv 1 \pmod 4 \implies \left(\frac{p-1}{2}\right)^2!\equiv -1 \pmod p$). Hence \begin{align} \frac{1\cdot 3\cdots (2k-1)}{2\cdot 4 \cdots 2k} \equiv \frac{1\cdot 3\cdots (p-2(k+1))}{2\cdot 4 \cdots (p-(2(k+1)-1))} \pmod p \end{align} then $a_k \equiv a_{\frac{p-1}{2}-k} \pmod p$ and the sum collapses a bit, \begin{align*} \sum_{k =1}^{\frac{p-3}{2}} \frac{a_k}{2k+1} \equiv \sum_{k =1}^{\frac{p-3}{4}} \frac{a_k}{2k+1} - \sum_{k =1}^{\frac{p-3}{4}} \frac{a_{\frac{p-1}{2}-k}}{2k} \equiv -\sum_{k =1}^{\frac{p-3}{4}} \left(\frac{a_k}{(2k)(2k+1)} \right) \pmod p \end{align*} but this doesn't seem to help. The symmetry of $a_k$ described above gives $\sum_{k =1}^{\frac{p-3}{2}}(-1)^k a_k \equiv 0 \pmod p$, which may be of interest to others.