Law of large numbers for triangular arrays whose moments "look independent"

Question

Let $(X_{n,k})_{k=1,\ldots,n}^{n\in\mathbb{N}}$ be a triangular array of random variables with finite moments of all orders, with no assumptions on their independence. Suppose that $$ \mathbb{E}\left[\frac1n \sum_{k=1}^n X_{n,k} \right] \xrightarrow{n\to\infty} \mu $$ If we further know that $$ \mathbb{E}\left[\left(\frac1n \sum_{k=1}^n X_{n,k}\right)^m \right] \xrightarrow{n\to\infty} \mu^m $$ for all $m \in \mathbb{N}$, is it possible to prove something more about $\frac1n \sum_{k=1}^n X_{n,k}$, for instance that it converges to $\mu$ in probability?

The question seems related to exchangeability. Nothing is lost here by assuming the rows $X_{n,1}\ldots,X_{n,n}$ are exchangeable, such that $$ (X_{n,1}\ldots,X_{n,n}) \stackrel{d}{=}(X_{n,\sigma(1)}\ldots,X_{n,\sigma(n)}) $$ for any permutation $\sigma \in \text{Perm}(n)$. In this case there are $o(n^m)$ terms in $\left(\sum_{k=1}^n X_{n,k}\right)^m$ which have non-distinct $k$. Then expression is dominated by terms which are distinct, so $$ \mathbb{E}\left[\left(\frac1n \sum_{k=1}^n X_{n,k}\right)^m \right] \approx \mathbb{E}[X_{n,1}X_{n,2}\cdots X_{n,m}] = \mathbb{E}[X_{n,1}]^m $$ If $X_{n,k}$ are i.i.d. then the condition is met and we have the law of large numbers. On the other extreme, if all $X_{n,k} = X$ for some random variable $X$, then the condition fails $$ \mathbb{E}\left[\left(\frac1n \sum_{k=1}^n X_{n,k}\right)^m \right] \xrightarrow{n\to\infty} \mathbb{E}[X^m] $$ and $\frac1n \sum_{k=1}^n X_{n,k} = X$, which does not converge to $\mu$ in any way. Does the condition imply that we are somehow more like the first example than the second?

Answer 1

$\newcommand\ep\varepsilon$Let $$Z_n:=\frac1n \sum_{k=1}^n X_{n,k}. \tag{1}\label{1}$$ We have $EZ_n\to\mu$ and $EZ_n^2\to\mu^2$, whence $\operatorname{Var}\,Z_n=EZ_n^2-(EZ_n)^2\to0$. So, for each real $\ep>0$, by Chebyshev's inequality, $$P(|Z_n-EZ_n|>\ep)\le\frac{\operatorname{Var}\,Z_n}{\ep^2}\to0.$$ So, $Z_n-EZ_n\to0$ in probability. Since $EZ_n\to\mu$, we conclude that $Z_n\to\mu$ in probability, as desired.

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As seen from here, the structure \eqref{1} of the random variables $Z_n$ plays no role. Also, the condition $$ E\Big(\frac1n \sum_{k=1}^n X_{n,k}\Big)^m \to \mu^m $$ for $m\ge3$ was not used or needed.