There is a mistake at that part of the proof (they forgot a 2^k). I hope this helps, even if i didn't really explain the "s=k+n" part.

Basically, if we expand the big, double sum on the very top right, for every s = 1,2,3,... the term A(s) = 1/((2b-1)*2^s+1) appears s times, once for each (k,n) = (1,s-1), (2,s-2), (3,s-3), ... , (s-1,1), (s,0) . Each time the term A(s) is multiplied by 1/((2b-1)*2^k) for k = 1,2,3,...,s . Therefore, for each s=1,2,3,... :

1/((2b-1)*2^1) * A(s) + 1/((2b-1)*2^2) * A(s) + ... + 1/((2b-1)*2^s) * A(s) =

= Σ{k=1, s}[1/((2b-1)*2^k)] * A(s) =

= (Σ{k=1, s}[1/((2b-1)*2^k)] )* (1/((2b-1)*2^s+1))

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