It is not very important, but since you mentioned it :
The interval of convergence of the Taylor series of 1/(1-z) at z=0 is indeed (-1,1).
But “1/(1-z) = 1 + z + O(z^2) for all z” does not make sense to me.
1/(1-z) = 1 + z + O(z^2) means that there is an M such as |1/(1-z) - (1 + z)| is no greater that M*z^2 for every z close enough to 0. It is about the behavior of 1/(1-z) - (1 + z) when z tends toward 0, not when z belongs to (-1,1).
I could not figure out why alpha > 0 neither and it seems wrong to me too. But this does not look like a problem.
We know that J is an increasing function because of 2-49. So in 2-53, alpha and log(x/S(x)) must have the same sign, since the remaining of the right member tends toward 0 when q tends toward + infinity.
Then b is positive and I think it is all that matters.
However, if alpha = 0, b is not defined. But if alpha=0 then log(x/S(x))=0 as a consequence of 2-53, so x/S(x)=1. There is only one x that gives us this since S is strictly decreasing. And by continuity we can still get 2-56.