This is a question originated from probability theory and economics. Probability measure induced on $\mathbb{N_{+}}$ is such that $P(\lbrace i \rbrace) = \frac{1}{2^i}$. Given random variable $X$: $(\mathbb{N_{+}}, F, P) \to (\mathbb{R},B(\mathbb{R}))$ such that $X(i) = 2^i$, it's easy too see that $X$ is not in $L^1$. But some increasing concave functions composited with $X$ is, for example, $log(X)$ and $\sqrt{X}$. Such functions are called utility functions, and in economics a utility function is always assumed to be concave and increasing. It's obvious that if the function is only concave, the expectation may also be infinite, like $F(x) = x$. But what about $\textbf{strictly concave and increasing functions?}$
So I want to ask : Will any increasing, strictly concave function $F: \mathbb{R_{++}} \to \mathbb{R}$ satisfy the condition that $\mathbb{E}(F(X)) = \sum\limits_{n=1}^{\infty}\frac{F(2^n)}{2^n} < \infty$?
My first thought is to find a possible upper bound for the growth rate of a strictly concave function, since we know that convex function can be expressed as the convex envelope: $f(x) = \sup_{(a,b)}ax+b$. So we also have concave envelope which is the infimum of all affine functions. Notice that strictly concave may imply that such concave function $F$ can never reach a 'straight line', so a possible guess is that $F(x) \leq x^{\alpha}$ for $\alpha \in (0,1)$ when $x$ large. If such guess is true, the question is proved. But I'm stuck here.
I'm wondering if my intuition is right, and if we had any results related to this. Could anyone prove or disprove the question that I asked? Thanks in advance for any help!
