We can solve this by back substitution method.
T(n) = T(n-1)+1/n
T(n-1) = T(n-2) + 1/(n-1)
T(n-2) = T(n-3) + 1/(n-2)
........
After moving like this, final equation we get as:
T(n) = 1+1/2+1/3+1/4+1/5+.......+1/(n-2)+1/(n-1)+1/n
T(n) = logn
So A is the correct answer.