Ok so here is the problem, basically it gives $\displaystyle 1 - 1/2 + 1/3 - .... - 1/(p-1) = a /(p-1)!$ and we have to show that $\displaystyle a =((2- 2^p)/p) modp.$
My attempt, I tried to convert the series in to:
$\displaystyle 1-1/2 + 1/3 -1/4 +... - 1/(p-1)= a/(p-1)!$
$\displaystyle 1+ 1/2 + 1/3+ ...+1/(p-1) = \sigma_{p-2}/(p-1)!$ from the Wolstenholmes congruence.


Then I added the latter and the former series to get:

$\displaystyle 2+ 2/3 + 2/5 + .... = a/(p-1)! + \sigma_{p-2}/(p-1)!$

then I tried to equate this with the given modular condition of a, and tried to play around with the equality but I just cant get it to be divisible by p or (modp).

Can anyone please give me some hints or a possible solution. It will be greatly appreciated.

Thank You