MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $X$:

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

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?

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.