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Given a sequence of positive numbers $X = (x_1, x_2, \dots)$, define the $k$th elementary symmetric mean of the first $n$ entries to be $$ S(X, n, k) := \frac{\displaystyle\sum_{1 \leq i_1 < \cdots < i_k \leq n} x_{i_1}x_{i_2}\cdots x_{i_k}}{\dbinom{n}{k}} $$

Note that $S(X,n,1)$ is the arithmetic mean and $S(X,n,n)^{1/n}$ is the geometric mean of the first $n$ entries of $X$.

Now we are going to impose the following conditions on our sequence $X$:

$$ \lim_{n \to \infty} S(X,n,n)^{1/n} = G < A = \lim_{n \to \infty} S(X, n, 1) < \infty$$

For any fixed $k$, I believe it's easy to show that these conditions imply

$$\lim_{n \to \infty} S(X,n,n-k)^{1/n-k} = G$$ and $$\lim_{n \to \infty} S(X,n,k)^{1/k} = A.$$

However, I'm interested in what happens when $k$ is allowed to depend on $n$. Specifically, is the existence of the AM and GM enough to imply the existence of

$$\lim_{n \to \infty} S(X,n,cn)^{1/cn} =: F(c)$$

where $c$ is any proportion in $(0,1)$? Maclaurin's inequalities tell us that the liminf/limsup lie inside [G,A], but do we also know these intermediate means don't oscillate strangely inside the interval?

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No. The number of terms matters only marginally in the regime $S(X,n,n/2)^{2/n}$ yielding a factor of $4$ at best when comparing everything to the largest product of $n/2$ elements to the power $2/n$ and the sum $X_1+\dots+X_{4}=4A$ and the product $X_1\dots X_{4}=G^4$ together with the order condition $X_1>X_2>X_3>X_4>0$ fail to determine $X_1X_2$ up to $4^2$ because for a fixed sum $2Q$ of $2$ positive numbers, the product can be any number in the interval $(0,Q^2]$, so choosing the sums $X_1+X_2=Q$ in two essentially different ways, we can then choose $X_3+X_4=2q$ in two ways with any desired ratio of products. Thus, you can get two 4-periodic sequences with the same $A$ and $G$, but different limits of $S(X,n,n/2)^{2/n}$. Mixing them in a sufficiently irregular way will result in no limit.

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