Answer
The series converges if ${p \gt 1}$ and diverges if ${p \le 1}$.
Work Step by Step
Write ${a_k} = f\left( k \right)$, where $f\left( x \right) = \dfrac{1}{{x\left( {\ln x} \right){{\left[ {\ln \left( {\ln x} \right)} \right]}^p}}}$.
Evaluate $\mathop \smallint \limits_3^\infty \dfrac{1}{{x\left( {\ln x} \right){{\left[ {\ln \left( {\ln x} \right)} \right]}^p}}}{\rm{d}}x$.
Let $t = \ln \left( {\ln x} \right)$. So, $dt = \dfrac{1}{{x\left( {\ln x} \right)}}dx$. The integral becomes
$\mathop \smallint \limits_3^\infty \dfrac{1}{{x\left( {\ln x} \right){{\left[ {\ln \left( {\ln x} \right)} \right]}^p}}}{\rm{d}}x = \mathop {\lim }\limits_{q \to \infty } \mathop \smallint \limits_{\ln \left( {\ln 3} \right)}^{\ln \left( {\ln q} \right)} \dfrac{1}{{{t^p}}}{\rm{d}}t = \mathop {\lim }\limits_{q \to \infty } \left[ {\dfrac{1}{{1 - p}}\left( {{t^{1 - p}}} \right)|_{\ln \left( {\ln 3} \right)}^{\ln \left( {\ln q} \right)}} \right]$
$ = \dfrac{1}{{1 - p}}\mathop {\lim }\limits_{q \to \infty } \left[ {{{\left( {\ln \left( {\ln q} \right)} \right)}^{1 - p}} - {{\left( {\ln \left( {\ln 3} \right)} \right)}^{1 - p}}} \right]$
Depending on the value of $p$, we obtain
$\mathop \smallint \limits_3^\infty \dfrac{1}{{x\left( {\ln x} \right){{\left[ {\ln \left( {\ln x} \right)} \right]}^p}}}{\rm{d}}x = \left\{ {\begin{array}{*{20}{c}}
{ - \dfrac{1}{{1 - p}}{{\left( {\ln \left( {\ln 3} \right)} \right)}^{1 - p}},}&{p \gt 1}\\
{\infty ,}&{p \le 1}
\end{array}} \right.$
Therefore, by the integral test, the series converges if ${p \gt 1}$ and diverges if ${p \le 1}$.
