Answer
$$
\sum_{n=2}^{\infty} \frac{1}{n\sqrt{ \ln n}}
$$
Apply the Integral Test to the given series we obtain
$$
\int_{2}^{\infty} \frac{1}{x\sqrt{ \ln x}} d x =\infty .
$$
So, the series diverges.
Work Step by Step
$$
\sum_{n=2}^{\infty} \frac{1}{n\sqrt{ \ln n}}
$$
Apply the Integral Test to the series
$$
\sum_{n=2}^{\infty} \frac{1}{n\sqrt{ \ln n}}
$$
The function
$$
f(x)=\frac{1}{x\sqrt{ \ln x}}
$$
is positive and continuous for $x\geq 2$ . To determine whether $f$ is decreasing, find the derivative.
$$
f^{'}(x)=-\frac{1+2\ln x}{2x^{2}(\ln x)^{\frac{3}{2}}}.
$$
So, $f^{'}(x) \lt 0 $ for $x\geq 2$,
and it follows that $f$ satisfies the conditions for the Integral Test.
You can integrate to obtain
$$
\begin{aligned}
\int_{2}^{\infty} \frac{1}{x\sqrt{ \ln x}} d x &=\lim\limits_{t \to \infty } \int_{2}^{t } \frac{1}{x\sqrt{ \ln x}} d x \\
& \quad\quad\quad\left[\text { use integration by substituting }
\right] \\
&\quad\quad\quad \left[ u=\ln x, \quad\quad d u= \frac{dx}{x }\right] , \text { then }\\
&=\lim\limits_{t \to \infty } \int_{2}^{t } \frac{1}{\sqrt{ u}} du \\
&=\lim\limits_{t \to \infty } [2 \sqrt{u}]_{2}^{t}\\
&=\lim\limits_{t \to \infty } [2 \sqrt{\ln x}]_{2}^{t} \\
&=\infty.
\end{aligned}
$$
So, the series diverges.