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.



Is there a simple expression for $\sum_{k =1}^{(p-3)/2} \frac{1\cdot 3\cdots (2k-1)}{2\cdot 4 \cdots 2k\cdot(2k+1)} \bmod p$?
It is known that $$\sum\_{k=0}^\infty\frac{\binom{2k}k}{(2k+1)4^k}=\frac{\pi}2.$$ Motivated by this, I proved in my paper [*On congruences related to central binomial coefficients*](https://doi.org/10.1016/j.jnt.2011.04.004) [J. Number Theory 131(2011), 2219-2238] the following result (as part (i) of Theorem 1.1 in the paper):


Let $p$ be any odd prime. Then
$$\sum\_{k=0}^{(p-3)/2}\frac{\binom{2k}k}{(2k+1)4^k}\equiv(-1)^{(p+1)/2}\frac{2^{p-1}-1}p\pmod{p^2}$$
and
$$\sum\_{k=(p+1)/2}^{p-1}\frac{\binom{2k}k}{(2k+1)4^k}\equiv pE\_{p-3}\pmod{p^2},$$
where $E\_0,E\_1,E\_2,\ldots$ are the Euler numbers.


