Numerical calculation suggests that for prime $p\ge 5$, \begin{align*} \sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}\sum_{i=\lfloor k/2\rfloor +1}^k\frac{1}{2i-1}\equiv 0\pmod{p}. \end{align*}

How can we arrive at this congruence?

Numerical calculation suggests that for prime $p\ge 5$, \begin{align*} \sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}\sum_{i=\lfloor k/2\rfloor +1}^k\frac{1}{2i-1}\equiv 0\pmod{p}. \end{align*}

Question. How can we prove this congruence?