= 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. Therefore
$$
\frac1{n(\log\log n)^r}>\frac1{n\log n}
$$
for all sufficiently large $n$. The comparison <series> diverges by the <integral test for convergence>, since
$$
\int^M\frac{dx}{x\log x}=\log\log M+\text{constant}\longrightarrow\infty.
$$
The <comparison test for series> gives
$$
\boxed{\sum_{n=3}^\infty\frac1{n(\log\log n)^r}\text{ diverges for every }r>0.}
$$
One 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.
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