Testing convergence of series $\sum_{n=3}^\infty\frac{1}{n (\ln(n))^p(\ln\ln(n))^q}$

We can useIntegral test for convergence. Thus we can write it as follows: $$I=\int _{ 3 }^{ +\infty }{ \frac { dx }{ x{ \left( \ln { x } \right) }^{ p }{ \left( \ln { \left( \ln { x } \right) } \right) }^{ q } } } =\int _{ \ln { 3 } }^{ +\infty }{ \frac { dt }{ { t }^{ p }{ \left( \ln { t } \right) }^{ q } } } $$ where we have substituted $t=\ln { x } $

If $p=1$ and $q>1$ then we get $$I=\int _{ \ln { \left( \ln { 3 } \right) } }^{ +\infty }{ \frac { dz }{ { z }^{ q } } } ={ \frac { { z }^{ -q+1 } }{ 1-q } }_{ \ln { \left( \ln { 3 } \right) } }^{ +\infty }<\infty $$ it means the integral converges.

Now consider that $p>1$ then for $\varepsilon >0$ and every arbitrary $\eta $ we know $\lim _{ t\rightarrow +\infty }{ \frac { { \left( \ln { t } \right) }^{ \eta } }{ { t }^{ \varepsilon } } } =0\quad $ we can write $\frac { 1 }{ { t }^{ p }{ \left( \ln { x } \right) }^{ q } } \le \frac { 1 }{ { t }^{ \alpha } } $ for big enough $t>0$ where $p\ge \alpha >1$

Similarly,if $p<1$ for big enough $t>0$ $\frac { 1 }{ { t }^{ p }{ \left( \ln { t } \right) }^{ q } } \ge \frac { 1 }{ { t }^{ \alpha } } $ inequation is true ,where $\quad p\le \alpha <1$ From all this we can consider this integral converges if $p>1$ and diverges if $p<1$ (both case $q$ is arbitrary)