How would you compare $O((\log n)^k)$ in relation to $O(n^c)$? How would you compare $O((\log n)^k)$ in relation to $O(n^c)$, $n,c \in \mathbb{N}*$ ? I'm very stuck on how to go about this.
I specifically need to see how $O((\log n)^{2021})$ relates to $O(n^3)$ and $O(\log n \cdot n^2)$. Thanks for any help I'm really stumped on this one. I tried using l'hôpital or proof by induction but I'm heading nowhere.
 A: You have found

*

*$\log(\log n) = o(\log n)\qquad$


*i.e. $\log n$ grows faster than $\log(\log n)$ at least for large enough $n$


*and so $2021 \log(\log n) = o(3 \log n)$


*and from this  $(\log n)^{2021} = o(n^3)$.
But large enough $n$ can require a big number.
That $2021$ makes a substantial difference, with $(\log n)^{2021} > n^3$ for $n$ between $3$ and about $2.75 \times 10^{2537}$.
For very large $n$, here from about $2.76 \times 10^{2537}$, $n^3> (\log n)^{2021}$ and increasingly so for larger $n$.
A: Here is an explicit bound
showing that
$\dfrac{(\ln(1+x))^k}{x^{c}}
\to 0$
as $x \to \infty$.
For $x > 0, c>0$
$\begin{array}\\
\ln(1+x)
&=\int_1^{1+x} \dfrac{dt}{t}\\
&=\int_0^{x} \dfrac{dt}{1+t}\\
&\lt\int_0^{x} \dfrac{dt}{(1+t)^{1-c}}\\
&=\int_0^{x} (1+t)^{c-1}dt\\
&=\dfrac{(1+t)^c}{c}|_0^{x}\\
&=\dfrac{(1+x)^c-1}{c}\\
&\lt\dfrac{(1+x)^c}{c}\\
&\text{so, replacing }c \text{ by } c/2\\
\ln(1+x)
&\lt\dfrac{(1+x)^{c/2}}{c/2}\\
&\text{or}\\
\dfrac{\ln(1+x)}{x^c}
&\lt\dfrac{2}{c(1+x)^{c/2}}\\
&\text{so that}\\
\dfrac{(\ln(1+x))^k}{x^{ck}}
&\lt\dfrac{2^k}{c^k(1+x)^{ck/2}}\\
&\text{Replacing } c \text{ by } c/k\\
\dfrac{(\ln(1+x))^k}{x^{c}}
&\lt\dfrac{2^k}{(c/k)^k(1+x)^{c/2}}\\
&=\dfrac{(2k/c)^k}{(1+x)^{c/2}}\\
\end{array}
$
