Convergence of $\sum_{n\ge16}^{ }\frac{1}{n\ln n\left(\ln\ln n\right)^{p}}$ How to show that $$\sum_{n\ge16}^{ }\frac{1}{n\ln n\left(\ln\ln n\right)^{p}}$$
Converges for $p>1$ Just by using Cauchy condensation test?

I applied the test , but I cannot continue:
$$\sum_{n\ge15}^{ }\frac{1}{n\ln2\left(\ln\left(n\ln2\right)\right)^{p}}$$
again:
$$\sum_{n\ge14}^{ }\frac{1}{\ln2\left(\ln\left(2^{n}\ln2\right)\right)^{p}}=\sum_{n\ge14}^{ }\frac{1}{\ln2\left(\ln\left(2^{n}\right)+\ln\ln\left(2\right)\right)^{p}}$$
again:
$$\sum_{n\ge13}^{ }\frac{1}{\ln2\left(2^{n}\ln\left(2\right)+\ln\ln\left(2\right)\right)^{p}}$$
So what now?
 A: The proper statement is here. Have a look before we proceed.
$$ \frac{2^n}{2^n\ln 2^n\left(\ln\ln (2^n)\right)^{p}}=\frac{1}{(n\ln 2)\left(\ln (n \ln 2)\right)^{p}}$$
Observe,
$$\lim_{n\to \infty}\dfrac{\frac{1}{n\left(\ln n\right)^{p}}}{\frac{1}{(n\ln 2)\left(\ln (n \ln 2)\right)^{p}}}=\lim_{n\to \infty}\dfrac{(n\ln 2)\left(\ln (n \ln 2)\right)^{p}}{n\left(\ln n\right)^{p}}=\ln 2$$
So series
$$\sum_{n\ge1}^{ }\frac{1}{(n\ln 2)\left(\ln (n \ln 2)\right)^{p}}$$ and $$ \sum_{n\ge1}^{ }\frac{1}{n\left(\ln n\right)^{p}} $$converges or diverges simultaneously. 
Now we look at a new series $$\sum_{n\ge1}^{ }\frac{1}{n\left(\ln n\right)^{p}}$$
Again we look at (Cauchy-condensation test)
$$\frac{2^n}{2^n\left(\ln (2^n)\right)^{p}}=\frac{1}{\left( n \ln (2)\right)^{p}}=\frac{1}{n^p\left(\ln (2)\right)^{p}}$$
Now we get the series $$\sum_{n\ge1}^{ }\frac{1}{n^p\left(\ln (2)\right)^{p}}$$
this last series converges when $p>1$. And hence by cauchy-condensation test the above series converges for  $p>1$.
You apply Cauchy-Condensation Test twice.
A: Applying the test once gives that the original series converges iff
$$
\sum\limits_{n = 15}^\infty  {\frac{1}{{n\log 2(\log n + \log\log 2)^p }}}
$$
converges. But
$$
\frac{2^p}{{\log 2}}\sum\limits_{n = 15}^\infty  {\frac{1}{{n\log ^p n}}} \geq \sum\limits_{n = 15}^\infty  {\frac{1}{{n\log 2(\log n + \log\log 2)^p }}}  \ge \frac{1}{{\log 2}}\sum\limits_{n = 15}^\infty  {\frac{1}{{n\log ^p n}}} .
$$
The series $\sum\nolimits_{n = 15}^\infty  {\frac{1}{{n\log ^p n}}}$ converges iff
$$
\frac{1}{{\log ^p 2}}\sum\limits_{n = 15}^\infty  {\frac{1}{{n^p }}} 
$$
converges, which happens precisely when $p>1$.
