Is there a simple expression for $\sum_{k =1}^{(p-3)/2} \frac{1\cdot 3\cdots (2k-1)}{2\cdot 4 \cdots 2k\cdot(2k+1)} \bmod p$?

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.

1 Answer

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 [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.

• Of course, $$\sum_{k=1}^{(p-3)/2}\frac{\binom{2k}k}{(2k+1)4^k}=\sum_{k=0}^{(p-3)/2}\frac{\binom{2k}k}{(2k+1)4^k}-1.$$ Aug 10, 2022 at 1:00
• The result indicates that it is natural to let $k$ start from $0$ rather than from $1$. Please note the infinite series for $\pi/2$. Aug 10, 2022 at 1:38
• Note that a prime such that $\tfrac{2^{p-1}-1}{p} \equiv 0 \bmod p$ is called a Wieferich prime en.wikipedia.org/wiki/Wieferich_prime . Only two such are known, 1093 and 3511, and there are very few results about them. Aug 10, 2022 at 3:45
• Note that the two congruences stated above already have the modulus $p^2$. Aug 10, 2022 at 22:15