Is this true: $(n \log n)^2 \in \Theta(\frac{n^3}{\log n})$

Question

I have to decide whether $(n \log n)^2 \in \Theta(\frac{n^3}{\log n})$ is true or false.

I can't really find a good approach, I'd assume it's wrong, but I can't prove it.

Consider $\frac{f(n)}{g(n)}$ as $n \rightarrow \infty$ where $f(n) = (n\log n)^2$ and $g(n) = \frac{n^3}{\log n}$ — fwd, Mar 12 '21 at 16:32

score 4 · Accepted Answer · edited Mar 12 '21 at 16:42

4

if $(n \log n)^2 \in \Theta(\frac{n^3}{\log n})$ there is a constant $C>0$ such that $$n^2\log^2n\ge \frac{Cn^3}{\log n}$$ for $n$ sufficiently large. This simplifies to $$\log^3n\geq Cn$$ which is false.

edited Mar 12 '21 at 16:42

zwim

28,563

answered Mar 12 '21 at 16:36

saulspatz

53,131

Is this true: $(n \log n)^2 \in \Theta(\frac{n^3}{\log n})$

1 Answers1