Take any non-negative iid random variables $X_1,X_2,...$.
Let $X^{(n)}=\max\{X_1,...,X_n\}$ be the highest order statistic among $n$ samples. Consider the expecetaction $E[X^{(n)}]$ as a function of $n$. Do we have restrictions on $E[X^{(n)}]$ except for that it is increasing and concave in $n$? In other words, for any function $f:\mathbb{N}\to \mathbb{R}_+$ which is increasing and concave, can we find a random variable such that $f(n)=E[X^{(n)}]$ for all $n$?
Here, $f$ being concave means $f(n+2)-f(n+1) \le f(n+1)-f(n)$ for any $n \in \mathbb{N}\cup\{0\}$.
If the CDF of $X$ is $F(x)$, the CDF of $X^{(n)}$ is $F(x)^n$. Assuming these expectations are finite, we have
$$ f(n) = \mathbb E[X^{(n)}] = \int_0^\infty \left(1 - F(x)^n\right)\; dx $$
But there's no need to restrict this formula to integer $n$: for any $t > 0$, let's take
$$ f(t) = \int_0^\infty \left(1 - F(x)^t\right)\; dx $$
Now the $k$'th derivative of this is
$$f^{(k)}(t) = \int_0^\infty - F(x)^t \ln(F(x))^k\; dx $$
which is nonnegative for odd $k$ and nonpositive for even $k$ (since $0 \le F \le 1$). As a consequence, the $k$'th differences of $f(n)$ are also nonnegative for odd $k$ and nonpositive for even $k$. So there are indeed other constraints on $f$.