 Solution

For any fixed $r>0$, $\log y=o(y^{1/r})$ as $y\to \infty$. For example, l'Hôpital's rule applied to $\log y/y^{1/r}$ gives the limit zero. With $y=\log n$, this implies $(\log \log n{)}^r<\log n$ eventually. Thereforefor all sufficiently large $n$. The comparison series diverges by the integral test for convergence, sinceThe comparison test for series givesOne may obtain the same growth bound by extending the given logarithmic estimate to real $y$ through $\lceil y\rceil$ and choosing any exponent smaller than $1/r$; no special convergence test for iterated logarithms is needed.