We have by the integral test:
$\sum_{n=2}^{\infty}\frac{1}{n}=\infty$
$\sum_{n=2}^{\infty}\frac{1}{n\ln n }=\infty$
$\sum_{n=2}^{\infty}\frac{1}{n\ln n (\ln\ln n) }=\infty$
$\sum_{n=2}^{\infty}\frac{1}{n\ln n (\ln\ln n) (\ln\ln \ln n)}=\infty$
and so on
Let
$a_1=n$
$a_2=n\ln n$
$a_3=n\ln n (\ln\ln n)$
$a_4=n\ln n (\ln\ln n) (\ln\ln \ln n)$
and so on
Then is
$\sum_{n=2}^{\infty}\frac{1}{a_n}$
diverging or converging?
Assuming that you mean $a_1 = 1$, $a_2 = 2\ln 2$, $a_3 = 3(\ln 3)(\ln \ln 3)$, and so on, the series does not even exist. $\ln \ln \ln 5 < 0$, so $\ln \ln \ln \ln 5$ will cause you a problem. (I suppose you could pick a branch of the complex $\ln$ function, but I'm pretty sure you won't get the series to converge in that case either.)