Answer
$$
\sum_{n=2}^{\infty} \frac{\ln n}{\sqrt n}
$$
Apply the Integral Test to the given series :
$$
\int_{2}^{\infty} \frac{\ln x}{\sqrt{x}} d x =\infty
$$
So, the series diverges.
Work Step by Step
$$
\sum_{n=2}^{\infty} \frac{\ln n}{\sqrt n}= \frac{\ln 2}{\sqrt 2}+ \frac{\ln 3}{\sqrt 3}+ \frac{\ln 4}{\sqrt 4}+...
$$
Apply the Integral Test to the series
$$
\sum_{n=2}^{\infty} \frac{\ln n}{\sqrt n}
$$
The function
$$
f(x)=\frac{\ln x}{\sqrt x}
$$
is positive and continuous for $x\geq2$ . To determine whether $f$ is decreasing, find the derivative.
$$
f^{'}(x)=\frac{\ln x}{\sqrt x}=\frac{2-\ln x}{2x^{\frac{3}{2}}}.
$$
So, $f^{'}(x) \lt 0 $ for $x\gt e^{2}$,
and it follows that $f$ satisfies the conditions for the Integral Test.
You can integrate to obtain
$$
\begin{aligned}
\int_{2}^{\infty} \frac{\ln x}{\sqrt{x}} d x &=\lim\limits_{t \to \infty } \int_{2}^{t } \frac{\ln x}{\sqrt{x}} d x \\
& \quad\quad\quad\left[\text { use integration by parts with }
\right] \\
&\quad\quad\quad \left[\begin{array}{c}{u=\ln x, \quad\quad dv= \frac{dx}{ \sqrt x } }\\ {d u= \frac{dx}{x }, \quad\quad v= 2\sqrt x }\end{array}\right] , \text { then }\\
&=\lim\limits_{t \to \infty } [2 \sqrt{x}(\ln x-2)]_{2}^{t}=\infty
\end{aligned}
$$
So, the series diverges.