Let $g$ be some function and $X$ a random variable. Assume $$ g_n (y) \sim h(y) \, n^{-1/2},$$ where $h(0)=1$ and $\sim$ stands for asymptotic equivalence ($\lim_n \frac{g_n(y)}{ h(y) n^{-1/2}} = 1$). Further, there exists a constant $c> 0$ such that for all $y$ and $n \in \mathbb{N}$ $$ g_n(y) \leq c\, h(y) \, n^{-1/2}.$$ Let $k \in \mathbb{N}$. I am asking myself whether there is a way to use the dominated convergence theorem so that $$ \lim_{n \to \infty} E\left( \frac{g_{n-k} (X) } {g_n (0) } \right) = E \left( \lim_{n \to \infty} \frac{g_{n-k} (X) } {g_n (0) } \right) = E (X).$$
I mean, I could use the upper bound for the numerator, but what to do about the denominator? The denominator is also deterministic whereas the numerator is random, but I can not just pull the denominator out of the expectation and use DCT, can I?
Assuming $g_n$ is non-negative and $h(X)$ has finite first moment, then yes. The trick is to multiply both numerator and denominator by $n^{1/2}$, then pull the (deterministic) denominator out. Specifically, we have
$$ E\left(\frac{g_{n-k}(X)}{g_n(0)}\right) = \frac1{n^{1/2}g_n(0)}E\Big(n^{1/2}g_{n-k}(X)\Big).$$
The term outside the expectation converges to $1$ as $n\to\infty$. On the other hand, we have
$$n^{1/2}g_{n-k}(X) \to h(X)$$
and
$$ 0 \le n^{1/2}g_{n-k}(X) \le c\left(\frac{n}{n-k}\right)^{1/2}h(X) \le c\sqrt{2} h(X) $$
for all $n \ge 2k$, and this right hand side is integrable. Thus, applying the dominated convergence theorem we find $$\lim_{n\to\infty}E\Big(n^{1/2}g_{n-k}(X)\Big) = E\big(h(X)\big),$$
and hence
$$\lim_{n\to\infty}E\left(\frac{g_{n-k}(X)}{g_n(0)}\right) = \lim_{n\to\infty}\frac1{n^{1/2}g_n(0)}\cdot\lim_{n\to\infty}E\Big(n^{1/2}g_{n-k}(X)\Big) = E\big(h(X)\big).$$