Supposing that #p > 3# with p prime, then
#1+1/2+1/3+...+1/(p-1)# has an even number of terms so we can establish an equivalent sum of terms
#1/k + 1/(p-k) = p/((p-k)k)#
then
#1+1/2+1/3+...+1/(p-1)=sum_(k=1)^((p-1)/2) p/((p-k)k)# so finally
#sum_(k=1)^(p-1)1/k equiv 0 mod p#
Now considering #sum_(k=1)^((p-1)/2) 1/((p-k)k)# we have
#(p-k)k equiv -k^2 mod p#.
After reducing the sum to an unique fraction, in the numerator will appear #(p-k)k, (k=1, cdots, (p-1)/2)# as factor into the #(p-1)/2# terms,so,
#sum_(k=1)^((p-1)/2) 1/((p-k)k)equiv-sum_(k=1)^((p-1)/2)k^2
mod p#
but #sum_(k=1)^((p-1)/2)k^2=((p-1)/2((p-1)/2+1)(2((p-1)/2)+1))/6 equiv 0 mod p#
So finally
#sum_(k=1)^(p-1)1/k equiv 0 mod p^2#
