Is O((n^2)*log(n)) greater than O(n^(2.5))?
In order to compare 2 complexities just calculate a limit of their ratios as below:

$\displaystyle\begin{align*}
 \lim_{n\to\infty}\frac{n^2log(n)}{n^2\sqrt{n}} 
   &amp;= \lim_{n\to\infty}\frac{log(n)}{\sqrt{n}} 
    = \lim_{n\to\infty}\frac{log(\sqrt{n})^2}{\sqrt{n}}
    = \lim_{n\to\infty}\frac{2log(\sqrt{n})}{\sqrt{n}} \\
    &amp;\underset{\left| k = \sqrt{n} \right|}{=}
      \ \ \lim_{k\to\infty}\frac{2log{(k)}}{k}
    = 2\lim_{k\to\infty}\frac{log{(k)}}{k} \\
    &amp;\leq 2\lim_{k\to\infty}\frac{ln{(k)}}{k} \\
    &amp;\overset{\ast}{=} 2\lim_{k\to\infty}\frac{(ln{(k))'}}{k'}
    = 2\lim_{k\to\infty}\frac{\frac{1}{k}}{1}
    = 2\lim_{k\to\infty}\frac{1}{k} \\
   &amp;= 0
\end{align*}$

We use L'Hôpital's rule to simplify calculating a limit for $\frac{\ln(k)}{k}$ at *.

As you can see, $O(n^2\times\log(n))$ is lower than the other.