Prove $\sum\limits_{n=2}^\infty \frac {1} {n\ln(n)\ln\ln{n}}$ is divergent. 
Evaluate if the following series is convergent or divergent: $\sum\limits_{n=2}^\infty \frac {1} {n\ln(n)\ln\ln{n}}$.

I cannot properly understand the notation that the book employs here $\ln\ln(x)$, but I guess it is referring to $\ln(\ln(x))$. Assuming $\ln \ln(x)=\ln(\ln(x))$,
I used the Weierstrass's or comparasion test to evaluate the series:
$$\sum\limits_{n=2}^\infty \frac {1} {n\ln(n)\ln\ln{n}}<\sum\limits_{n=2}^\infty \frac {1} {n\ln(n)\ln(n)}=\sum\limits_{n=2}^\infty \frac {1} {n\ln^2{n}}<\sum_\limits{n=2}^{\infty}\frac{1}{n^2},$$ proving the series converge.
However the solution point out the series diverge. 
I have already proved the convergence $\sum_\limits{n=2}^{\infty}\frac{1}{n^2}$.
Question:
What am I doing wrong? How can I prove the series diverge? Is $\ln\ln(x)=\ln(\ln(x))$ meant by the author of the book?
 A: Your series diverges by the integral test, because$$\int\frac1{x\log(x)\log\bigl(\log(x)\bigr)}\,\mathrm dx=\log\bigl(\log\bigl(\log(x)\bigr)\bigr)$$and because $\lim_{x\to+\infty}\log\bigl(\log\bigl(\log(x)\bigr)\bigr)=+\infty$.
Concerning your approach, note that the inequality $\frac1{n\ln^2n}<\frac1{n^2}$ is false. Actually, $n\ln^2n<n^2$.
A: HINT
Let use Cauchy condensation test
$$ 0 \ \leq\ \sum_{n=1}^{\infty} f(n)\ \leq\ \sum_{n=0}^{\infty} 2^{n}f(2^{n})\ \leq\ 2\sum_{n=1}^{\infty} f(n)$$
$$\sum\limits_{n=2}^\infty \frac {1} {n\ln(n)\ln\ln{n}}\ge \frac12\sum\limits_{n=2}^\infty \frac {2^n} {2^n\ln(2^n)\ln\ln{2^n}}=\frac12\sum\limits_{n=2}^\infty \frac {1} {n\ln 2\ln(n\ln 2)}$$
and
$$\sum\limits_{n=2}^\infty \frac {1} {n\ln 2\ln(n\ln 2)}\ge \frac12 \sum\limits_{n=2}^\infty \frac {2^n} {2^n\ln 2\ln(2^n\ln 2)}=\frac12 \sum\limits_{n=2}^\infty \frac {1} {\ln 2\cdot (n\ln 2+\ln \ln 2)}$$
A: Another way is to use Cauchy condensation test. It is usually a good option when there are logarithms. You have to use it twice here. 
A: Alternatively, use Ermakoff's test:
$$\lim_\limits{x\to\infty} \frac{e^xf(e^x)}{f(x)}=\lim_\limits{x\to\infty} \frac{e^x\cdot \frac{1}{e^x\cdot x\cdot \ln x}}{\frac{1}{x\cdot \ln x\cdot \ln \ln x}}=$$
$$\lim_\limits{x\to\infty} \ln \ln x=\infty.$$
Hence, the series diverges.
