Test the convergence of the series $$\sum_{n=2}^{\infty}{\frac{1}{\sqrt{n}\ \log(n)}}$$
Using the comparison test $\log(n)<\sqrt{n}$ for all $n \in N$. From this it follow that $\frac{1}{\sqrt{n}\ \log(n)}>\frac{1}{\sqrt{n}\sqrt{n}}=\frac{1}{n}$ which is a diverging. So our original series also diverges.
On
It looks great to me. But, alternatively, you could apply Cauchy's Condensation Test, which states that
$\sum a_n$ converges if and only if $\sum 2^na_{2^n}$ converges. In this case, the general term is:
$a_n=\frac{1}{\sqrt{n} \log n}$. Then $2^na_{2^n} = \frac{2^n}{\sqrt{2^n}\log2^n} = \frac{1}{\log2}\frac{\sqrt{2^n}}{n}$. Therefore, it all boils down to the convergence of $\sum \frac{\sqrt{2^n}}{n}$. But since $\frac{\sqrt{2^n}}{n} > \frac{1}{n}$ for $n\geq1$, this series diverges.
Hence, the original series diverges, just as you correctly proved.
A comment, but too long for comment.
To bound $\ln(x)$ for $x > 1$, for any $c > 0$,
$\begin{array}\\ \ln(x) &=\int_1^x \dfrac{dt}{t}\\ &=\int_1^x t^{-1}dt\\ &<\int_1^x t^{-1+c}dt\\ &<\dfrac{t^{c}}{c}\big|_1^x \\ &<\dfrac{x^{c}-1}{c}\\ &<\dfrac{x^{c}}{c}\\ \end{array} $
For example, if $c = \frac12$, this gives $\ln(x) \lt 2\sqrt{x}$.
By setting $c=a/2$, $\ln(x) \lt \dfrac{x^{a/2}}{a/2} $ or $\dfrac{\ln(x)}{x^a} \lt \dfrac{2x^{-a/2}}{a} $ so $\lim_{x \to \infty}\dfrac{\ln(x)}{x^a} = 0 $ for any $a > 0$.