Does $20n$ belong to $O(n^{1-\epsilon})$ for some $\epsilon > 0$?
Big O notation hides constant factors. In particular, for every constant $C &gt; 0$,
$$
f \in O(g) \Longleftrightarrow Cf \in O(g).
$$
In particular, for any $\epsilon &gt; 0$,
$$
20n \in O(n^{1-\epsilon}) \Longleftrightarrow n \in O(n^{1-\epsilon}).
$$
This follows directly from the definition of big O. Indeed, suppose that $f \in O(g)$. Then there exist $N,A&gt;0$ such that $f(n) \leq Ag(n)$ for all $n \geq N$. This implies that $Cf(n) \leq ACg(n)$ for all $n \geq N$, and so $Cf = O(g)$. Similarly, if $Cf = O(g)$ then the same argument shows that $f = C^{-1} C f = O(g)$.