Is my solution for the divergence of this series correct? 
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.
 A: 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: 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 $a > 0$.
