Let s be a point between sc and p-r. All the points in the disk, centered at p, produce convergent series, as demonstrated by abel summation. As a bonus, the speed of convergence is determined by the (real) distance from s. The left edge of the disk, at p-r, has the slowest convergence. Therefore, convergence is uniform across the entire disk.
Since zeta is an infinite sum of analytic functions, and the series converges uniformly, zeta is analytic. Its derivative is the sum of the derivatives of the constituent functions fn/ns. this holds for any point p to the right of sc, hence ζ(f,s) is analytic on the half plane to the right of sc.
The derivatives of zeta can be characterized. Concentrate on a term fn/ns. Of course fn is a constant, so find the derivative of n-s, with respect to s. Rewrite this expression as E-log(n)s. The derivative is now -log(n) times the original function.
Put this all together and ζ(f,s)′ = -ζ(log(n).f,s). Repeat this k times to find the kth derivative of ζ(f,s). Each fn is multiplied by -log(n) to the k.
Did we lose absolute convergence? Not at all. The terms of zeta, to the right of sa, are bounded by some constant times rn for r < 1. Replace rn with log(n)rn, and apply the ratio test. For large n, the ratio approaches r. Let r′ be slightly larger than r, and still less than 1. The norms of the terms are bounded by r′n, and the derivative is absolutely convergent. In other words, sa cannot move to the right.
Suppose sa moves to the left. Let s be a point that has become absolutely convergent under ζ′. Its terms are bounded by rn. Divide by log(n), and the original ζ(f,s) was absolutely convergent at s, which is a contradiction. Therefore sa does not move with differentiation.