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? 
 A: According to Rudin's criterion (or whatever its name is)
$$\sum_{n=3}^\infty \frac{1}{n \log n (\log \log n)^p}$$
converges iff
$$
\sum_{k=2}^\infty \frac{2^k}{2^k \log (2^k) (\log \log (2^k))^p}
$$
converges or equivalently if
$$
\sum_{k=2}^\infty \frac{1}{k\log 2 (\log (k\log 2))^p}=
\frac{1}{\log 2}\sum_{k=2}^\infty \frac{1}{k (\log k+\log\log 2)^p}
$$
converges.
But
$$
\log k\ge \log k+\log\log 2\ge \frac{1}{3}\log k,
$$
and hence the sum
$$
\frac{1}{\log 2}\sum_{k=2}^\infty \frac{1}{k (\log k+\log\log 2)^p}$$
converges iff
$$
\frac{1}{\log 2}\sum_{k=2}^\infty \frac{1}{k (\log k)^p}
$$
converges, which we already know that it converges iff $p>1$.
A: 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?
A: Let $p$ be any given positive real number. Then, for each $n \in \mathbb{N}$ such that $n \geq 3$, we have 
$$ \frac{ 1 }{ n \ln n (\ln \ln n )^p } > \frac{ 1 }{ (n+1) \ln (n+1) (\ln \ln (n+1) )^p } > 0. $$
Thus the series 
$$ \sum_{n=3}^\infty \frac{1}{n \ln n (\ln \ln n)^p}$$
converges if and only if the series 
$$ \sum_{n=4}^\infty \frac{1}{n \ln n (\ln \ln n)^p}$$
converges, and the latter series converges if and only if the series 
$$ \sum_{k=2}^\infty \frac{2^k}{2^k \ln 2^k \left( \ln \ln 2^k \right)^p} =  \sum_{k=2}^\infty \frac{1}{ k (\ln 2) \  ( \ln k + \ln \ln 2 )^p } $$ converges, and the last series converges if and only if the series 
$$ \sum_{k=2}^\infty \frac{1}{ k ( \ln k + \ln \ln 2 )^p } \tag{A} $$ converges.
Now we have the following result: 

Let $\sum x_n$ and $\sum y_n$ be series of positive real numbers such that the limit $$ \lim_{n \to \infty} \frac{ x_n }{y_n} $$ exists in the extended real number system. Then the following three statements hold. 
(a) Suppose that $$ 0 <  \lim_{n \to \infty} \frac{ x_n }{y_n} < +\infty. $$ Then $\sum x_n$ converges if and only if $\sum y_n$ converges. 
(b) Suppose that $$ \lim_{n \to \infty} \frac{ x_n }{y_n} = 0. $$ If $\sum y_n$ converges, then so does $\sum x_n$. 
(c) Suppose that $$ \lim_{n \to \infty} \frac{ x_n }{y_n} = +\infty. $$ If $\sum x_n$ converges, then so does $\sum y_n$. 

Here is the  proof: 

Let us put $$ l \colon= \lim_{n \to \infty} \frac{ x_n }{y_n}. $$
(a) If $0 < r < +\infty$, then $0 < r/2 < +\infty$ as well, and thence for all sufficiently large $n$, we have 
  $$ \left\lvert \frac{ x_n }{y_n}  - r \right\rvert < \frac{r}{2},  $$
  which is equivalent to 
  $$ \frac{l}{2} < \frac{x_n}{y_n} < \frac{3l}{2}, $$
  which in turn implies 
  $$ \frac{l}{2} y_n < x_n < \frac{3l}{2} y_n $$
  for all sufficiently large $n$. Now we can apply Theorem 3.25 (a) in Rudin to arrive at our desired result. 
(b) If $l = 0$, then we have 
  $$ 0 <  \frac{x_n}{y_n}  =  \left\lvert \frac{x_n}{y_n} - 0 \right\rvert < 1,$$
  and hence $$ x_n < y_n, $$ 
  for all sufficiently large $n$, and we can again apply Theorem 3.25 (a) in Baby Rudin. 
(c) If $l = +\infty$, then we have 
  $$ \frac{x_n}{y_n} > 1, $$
  and hence $$ x_n > y_n, $$
  for all sufficiently large $n$, and again Theorem 3.25 (a) in Rudin leads to our desired conclusion. 

Now let us put 
$$ x_k \colon= \frac{1}{ k ( \ln k + \ln \ln 2 )^p }, $$
and 
$$ y_k \colon= \frac{1}{ k ( \ln k )^p } $$
for each $k \in \mathbb{N} $ such that $k \geq 2$. 
Then we note that 
$$ \frac{x_k}{y_k} = \frac{ \frac{1}{ k ( \ln k + \ln \ln 2 )^p }  }{ \frac{1}{ k ( \ln k )^p } } = \frac{ ( \ln k )^p  }{ ( \ln k + \ln \ln 2 )^p } = \frac{ 1 }{  \left( 1 + \frac{ \ln \ln 2 }{ \ln k } \right)^p },  $$
and thus 
$$ \lim_{k \to \infty} \frac{ x_k }{y_k} = \lim_{k \to \infty} \frac{ 1 }{  \left( 1 + \frac{ \ln \ln 2 }{ \ln k } \right)^p } = \frac{ 1 }{ (1 + 0)^p} = 1. $$
So the series $\sum x_n$  converges if and only if the series $\sum y_n$ converges. Now we know from Theorem 3.29 in Baby Rudin that the series $\sum y_n$ converges for $p > 1$ and diverges for $0 < p \leq 1$. Therefore, the series in (A) above converges for $p > 1$ and diverges for $0 < p \leq 1$. 
Hence our original series converges for $p > 1$ and diverges for $0 < p \leq 1$. 
As for $p = 0$, we note that the series 
$$ \sum_{n = 2}^\infty \frac{1}{n \lin n} $$
diverges, again by Theorem 3.29 in Rudin. 
Finally, for $p < 0$, we note that for sufficiently large $n$, 
$$ 0 < ( \ln \ln n )^p < 1, $$
and so 
$$ \frac{1 }{ ( \ln \ln n )^p } > 1, $$
which implies that 
$$  \frac{ 1 }{ n \ln n \ ( \ln \ln n )^p } > \frac{1 }{ n \ln n}. \tag{B} $$
Now as the series $$ \sum \frac{1}{n \ln n}$$ diverges by Theorem 3.29 in Rudin, so we can conclude from (B) and Theorem 3.25 (b) in Rudin that the series 
$$  \frac{ 1 }{ n \ln n \ ( \ln \ln n )^p } $$
diverges. 
Hence our original series converges for $p > 1$ and diverges for $p \leq 1$. 
