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?



Law of large numbers for triangular arrays whose moments "look independent"
$\newcommand\ep\varepsilon$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.


