calculus - How to decide about the convergence of $sum(nlog nloglog n)^{-1}$?

In Baby Rudin, Theorem 3.27 on page 61 reads the following:

Suppose $a_1 \geq a_2 \geq a_3 \geq \cdots \geq 0$. Then the seires $\sum_{n=1}^\infty a_n$ converges if and only if the series
$$\sum_{k=0}^\infty 2^k a_{2^k} = a_1 + 2a_2 + 4a_4 + 8a_8 + \ldots$$ converges.

Now using this result, Rudin gives Theorem 3.29 on page 62, which states that

If $p>1$, $$\sum_{n=2}^\infty \frac{1}{n (\log n)^p}$$ converges; if $p \leq 1$, the series diverges.

Right after the proof of Theorem 3.29, Rudin states on page 63:

This procedure may evidently be continued. For instance, $$\sum_{n=3}^\infty \frac{1}{n \log n \log \log n}$$ diverges, whereas $$\sum_{n=3}^\infty \frac{1}{n \log n (\log \log n)^2}$$ converges.

How do we derive these last two divergence and convergence conclusions
by continuing the above procedure as pointed out by Rudin?

I mean how to prove the convergence of the seires
$$\sum_{n=3}^\infty \frac{1}{n \log n (\log \log n)^2}?$$

And, how to prove the divergence of $$\sum_{n=3}^\infty \frac{1}{n \log n \log \log n}$$ using the line of argument suggested by Rudin?