How to decide about the convergence of $\sum(n\log n\log\log n)^{-1}$? 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? 

Using this method which's called Cauchy condensation we get
$$\sum_{k\ge1}\frac{2^k}{2^k\ln 2^k\ln\ln(2^k)}=\frac1{\ln2}\sum_{k\ge1}\frac1{k\ln(k\ln2)}\sim\frac1{\ln2}\sum_{k\ge1}\frac1{k\ln(k)}$$
so the series
$$\sum_{n\ge1}\frac1{n\ln n\ln(n\ln n)}$$
is divergent. Can you now solve the second series?
