Testing convergence of series $\sum_{n=3}^\infty\frac{1}{n (\ln(n))^p(\ln\ln(n))^q}$ Lets have this problem. $$\sum_{n=3}^\infty\frac{1}{n (\ln(n))^p(\ln\ln(n))^q}$$
I have rewritten this to a form
$$\sum\frac{1}{np'^{\ln\ln(n)}q'^{\ln\ln\ln(n)}}$$
For $p,q\in\mathbb{R}$. Obviously, $p',q'>0, p=\ln(p'),q=\ln(q')$
 A: 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)
A: Now I tried something. And it is not as crazy as the "duplicate" question. It is the classic Lobachev/Cauchy/Leibnitz condensation rule:
$$\sum a_n\,\text{converges}\iff\sum 2^na_{2^n}\,\text{converges}$$
so applying this
$$\sum\frac{1}{n(\ln(n))^p(\ln(\ln(n)))^q}\sim\sum\frac{2^n}{2^nn^p(\ln2)^p(\ln(n\ln2))^q}$$
now $(\ln2)^p$ and $\ln 2$ are constants, so I forget them(multiplication by finite constants does not change the convergence of the series). I have
$$\sum\frac{1}{n^p(\ln(n))^q}\sim\sum\frac{1}{2^{n(p-1)}n^q}$$
by applying the criterion the second time. But to test the convergence of the latter series, I need nothing more than an integral criterion.
$$\int_3^\infty n^{-q}e^{\ln(2)(n(p-1))}dn$$
by substituing $ln(2)(n(1-p))=u$ and $\frac{du}{ln(2)(1-p)}=dn$ we only need to find out what is $n^{-q}$. So $$n=\frac{u}{(1-p)\ln2}$$ and so $$n^{-q}=\left(\frac{u}{(1-p)\ln2}\right)^{-q}$$
and we have the integral
$$\left(\frac{1}{(1-p)\ln2}\right)^{-q+1}\int_3^\infty u^{-q}e^{-u}du\leq\left(\frac{1}{(1-p)\ln2}\right)^{1-q}\Gamma(1-q) $$
which makes sense for $q>1$ and $p>1 $. However, since the original integral does not start from zero, we can easily set $q=1$ also, then we don't need no gamma function, just evaluate the integral $$\int_3^\infty e^{-u}/u du<\infty$$
which converges to something positive because the limits in the bounds exist, are finite, the function is positive, monotone, continouous and goes to zero in infinity. But this is just the case $q=1,p>1$, else we can not substitute and have to integrate the fucntion $e^u/u$, which diverges. The case $p=1,q>1$ is integral
$$\int_3^\infty n^{-q}dn<\infty$$
by the $p$(in this case $q$-test). So the case $p=1,q>1$ also holds.
But I am not 100% sure if this is correct and not really sure if I am not missing some conditions by this. 
