Your first question has an obvious answer: The series $S$ diverges, because $$k^{\frac{1}{k}}-1 = \exp\left(\frac{\ln k}{k}\right)-1>\frac{\ln k}{k} > \frac{1}{k}$$ and $\sum_{k=2}^\infty 1/k=\infty$.
Carlo Beenakker
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