Asymptotic Growth: little o(n) versus $O(n^\alpha)$ Let $f(n) \geq 0$ be defined for all $n \in \mathbb{N}$. Suppose $f(n)$ is $o(n)$ and at the same time $f(n)$ is not $O(n^\alpha)$ for all $0 \leq \alpha < 1$. Is this necessarily a contradiction?
I'm using the following definitions:
$f(n)$ is $O(n^\alpha))$ if there exists constants $\alpha, c$ and $n_0$ such that for all $n \geq n_0$, $f(n) \leq c n^\alpha$.
$f(n)$ is $o(n)$ if $\lim_{n \rightarrow \infty} f(n)/n = 0$.
I don't think it's necessarily a contradiction since no matter how close $\alpha$ is to $1$ at some point the graph of $cn^\alpha$ will be below the graph of $n$, and so it would seem possible to construct a function satisfying both conditions.

You are right that this is not a contradiction. An example of a function of $n$ satisfying this is
$$\frac{n}{\log n}$$
