When the mean of a non-negative, integer-valued random variable goes to zero, does this imply anything about the other raw moments? When the mean of a non-negative, integer-valued random variable $X_n$, for example counting paths in a random graph on $n$ vertices between two fixed vertices, goes to zero,
$$\lim_{n \to \infty}\mathbb{E}[X_n] =0$$
can we have $\lim_{n \to \infty}\mathbb{E}[X_n^3] = \infty$, or under what conditions does this occur? This seems reasonable given we just need two different sequences to have different limits, but a bit counter-intuitive, since we're counting objects, and the raw moments give the typical number of ordered pairs, triples, etc.

Here is a concrete example about counting things in random graphs.
Let $H$ be the $5$-vertex graph obtained by adding an edge to $K_4$:

Suppose that $X_n$ counts the number of copies of $H$ in $G_{n,p}$, for some carefully chosen $p$. Then $\mathbb E[X_n] = \Theta(n^5 p^7)$. So if we choose $p = o(n^{-5/7})$, then $\mathbb E[X_n] \to 0$ as $n \to \infty$.
However, for such values of $p$, we have $\mathbb E[X_n^2] = \Theta(n^6 p^8)$: $X_n^2$ counts ordered pairs of copies of $H$, and the $\Theta(n^6 p^8)$ term comes from two copies that share $K_4$ but have different edges coming out of the $K_4$. Whenever $p$ is (asymptotically) between $n^{-3/4}$ and $n^{-5/7}$, such as $p = n^{-0.74}$, we have $\mathbb E[X_n] \to 0$ but $\mathbb E[X_n^2] \to \infty$ as $n \to \infty$.
Similarly, there is a $\Theta(n^7 p^9)$ contribution to $\mathbb E[X_n^3]$ that comes from three copies of $H$ that all share the same $K_4$. So if we take a value of $p$ such as $n^{-0.77}$, then we get $\mathbb E[X_n] \to 0$, $\mathbb E[X_n^2] \to 0$, and $\mathbb E[X_n] \to \infty$ as $n \to \infty$.
Essentially, when you see a situation where the mean goes to $0$ but higher moments go to infinity, this tells you that $X_n$ counts objects that rarely appear, but appear in bunches when they do. That's what we see here: it is rare to find a copy of $H$ in $G_{n,p}$ for this range of $p$, but if you find one copy, you find many copies.
