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?
> 
> 
> 



How to prove $\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}\sum_{i=\lfloor k/2\rfloor +1}^k\frac{1}{2i-1}\equiv 0\pmod{p}$?
It turned out the OP's question required a good amount of work and a joint effort (with Roberto Tauraso). However, "*the margin here is too small to contain the proof*", so to say.


Instead, readers can find the resolution [at this link](https://www.math.temple.edu/~tewodros/Harmonic_Sumsvnew.pdf). In addition, we prove that
\begin{align\*}
\sum\_{k=1}^{\infty}\frac{(-1)^k}{k}\sum\_{i=\lfloor k/2\rfloor +1}^k\frac{1}{2i-1}=-\frac58\zeta(2).
\end{align\*}


Comments and suggestions are welcome. 


