Is $\log{n}$ bounded from above by $n^{o(1)}$?

Question

Let $O(n)$ be "Big-O" of $n$ and $o(n)$ be "Small-O" of $n$.

It is a well-known fact that $O(n \log{n}) \subset O(n^{1 + \epsilon})$ for any $\epsilon > 0$. Can we omit the $\epsilon$, and just type $O(n \log{n}) \subset O(n^{1 + o(1)})$?

Answer 1

Consider the function $f(n) = n^{1 + \frac{1}{n}}$ which I guess you'd say is in "$O(n^{1 + o(1)})$". I'm not sure you can call it polynomial, though.

Now, compute

$\qquad \displaystyle \lim_{n \to \infty} \frac{n^{1 + \frac{1}{n}}}{n \log n} = 0$,

so you have in fact that $f \in o(n \log n)$. Therefore, functions in this funky class of yours are not even (all) asymptotic upper bounds for $n \log n$.

The difference is that $\varepsilon$ ensures a non-vanishing distance from $n^1$ whereas $o(1)$ means that functions of the form $n^{1 + o(1)}$ in fact converge towards $n \in o(n \log n)$.

Answer 2

Common usage would have $n\log n = n^{1 + o(1)}$. This is a shortcut for $n\log n = n^{1 + f(n)}$ for some function $f(n)$ satisfying $f(n) = o(1)$. In our case, this function is $f(n) = \log \log n / \log n$, which indeed satisfies $f(n) = o(1)$.