#### Answer

Divergent

#### Work Step by Step

Given: $\Sigma_{n =2}^{\infty} \frac{(-1)^{n}\sqrt n}{ln n}$
Check the derivative of the corresponding function.
$f(x)=\frac{\sqrt x}{ln x}$
$f'(x)=\frac{lnx-2}{2\sqrt x (lnx)^{2}}$
Find where $f'(x)=0$
$ln(x)-2=0$
$ln(x)=2$
$x=e^{2}\approx 7.4$
Test the interval after the critical number.
$(e^{2},\infty )$: $f'(8)\approx 0.003\gt 0$
It is positive as $x \to \infty$, so $f$ is decreasing for $x\gt e^{2}$.
Since, $\frac{\sqrt n}{ln n}$ is increasing.
$\lim\limits_{n \to \infty}\frac{(-1)^{n}\sqrt n}{ln n}=DNE$
The limit does not exist so the series diverges by the Test of Divergence.