Expectation of a transformed random variable I'm trying to prove the following:
Let $X_n$ be a sequence of positive random variables and $g$ be a positive function. Suppose that $E[X_n]\to \infty$ as $n\to\infty$. If $E[g(X_n)]$ exists, there exists a positive and finite constant $c$ such that 
$$
\lim_{n\to\infty} \frac{E[g(X_n)]}{g(E[X_n])} \to c.
$$
I don't know whether it is true or not. I tested some known random variables and I failed to find counter examples.
Here is one example: I select $g(x) = 1/x$ and $X_n\sim Gamma(2,n/2)$, where $Gamma(k,\theta)$ denotes the Gamma random variable of which density function is
$$
f(x)=\frac{1}{\Gamma(k)\theta^k}x^{k-1}e^{x/\theta}. 
$$
After some manipulation, I obtain $E[X_n] = n$ and $E[1/X_n] = 2/n $ so that the above statement is true with $c=2$. 
 A: The claim is false. Define $X_n$ to be
$$
X_n=\begin{cases}
\delta_n&\text{with probability $1-\epsilon_n$}\\
A_n&\text{with probability $\epsilon_n$}
\end{cases}
$$
where $\epsilon_n$ and $\delta_n$ are sequences of positive numbers tending to $0$, while $A_n$ is a sequence tending to $\infty$ such that $A_n\epsilon_n\to\infty$ (so that $E(X_n)\to\infty$). With appropriate choices of these sequences you can find $g(x)$ such that the limit in question is zero, or infinity.
Example: If $A_n=n^2$ and $\epsilon_n=\delta_n=\frac1n$, the limit is zero for $g(x):=x/(1+x)$; the limit is $\infty$ for $g(x):=x^2$.
The problem is that the requirement $Eg(X_n)<\infty$ is trivial to satisfy for these $X_n$; moreover, there are too many choices of $g$.
A: Another simple recipe for counterxamples: assuming $g(x)$ is differentiable we get:
$$E[g(X)]=g(\mu) + \frac{g^{''}(\mu)}{2!}m_2+ \frac{g^{'''}(\mu)}{3!} m_3+\cdots$$
where $\mu=E(X)$ and $m_i$ are the centered moments. Then, by choosing $g(x)=e^x$ we get
$$\frac{E[g(X)]}{g(E[X])}=1 + \frac{m_2}{2!}+ \frac{m_3}{3!} +\cdots$$
For example, let $X_n$ uniform on $[0,n]$ and $g(x)=e^x$
