i have some opinions about the second one

sum from k=2 to infinity a^(logb k)

It is equal to sum from k=2 to infinity k^(logb a)

= sum from k=2 to infinity k^-(logb 1/a)

= sum from k=1 to infinity k^-(logb 1/a) - 1

= Z[ logb(1/a) ] - 1 where Z is Riemann Zeta Function