This series is divergent.

By the conclusion of # 4538, we have $\sqrt{\ln{x}} > \ln{\ln n}$. Therefore

$$(\ln{n})^{\ln \ln n} = e^{{\ln \ln n}^{\ln \ln n}}=e^{(\ln \ln n)^2}<e^{\ln{n}}=n$$

Therefore, 

$$\sum_{n=2}^{\infty}\frac{1}{(\ln n)^{\ln \ln n}} > \sum_{n=2}^{\infty}\frac{1}{n}$$

It is well-known that $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n}$ diverges, therefore so will be the given series.