Rate of convegence of $\frac{E^2[X^n]}{ E[X^{n-1}] E[X^{n+1}]}$ as $n \to \infty$ Let $X$ be a non-negative discrete random variable such that $0\le X\le B$ and$$f(n)=\frac{E^2[X^n]}{E[X^{n-1}]E[X^{n+1}]}$$for $n \ge 1$.
I am interested in the rate of growth of $f(n)$ as $n \to \infty$. In other words, is there a way to quantify how fast $f(n)$ approaches its limit as a function of $n$ and some “property” of $X$?
A few observations:

*

*Upper bound: By Cauchy-Schwarz we have that $f(n) \le 1$.

*Exact limit: It can be shown that $\lim\limits_{n \to \infty} f(n)=1$. So, we are trying to refine this limit.

*Trivial example: If $X$ is supported on $\{0,B\}$, then  $f(n)= 1$ for all $n \ge 1$.

*Connection to Strong Cauchy-Schwarz: Essentially, this is a question about the sharper versions of Cauchy-Schwartz inequality. Therefore, using the expression for the correction term in Cauchy-Schwartz we have
$$
f(n)=1-\frac{1}{2} E \left[ \left| \frac{X^\frac{n-1}{2}}{ \sqrt{E[X^{n-1}]}}  -\frac{X^\frac{n+1}{2}}{ \sqrt{E[X^{n+1}]}}\right|^2 \right]. 
$$
I would like to answer this with minimal assumptions, but, if needed, we can assume there is a mass at $B$.

Some idea
Let $m\ge 2$. Let $0 \le a_1 < a_2 < \cdots < a_m$ and $\mathrm{Pr}(a_1) = p_1, \mathrm{Pr}(a_2) = p_2, \cdots, \mathrm{Pr}(a_m) = p_m$ with $p_1+p_2+\cdots + p_m = 1$. We have
$$f(n) = \frac{(1+A)^2}{(1+B)(1+C)}$$
where
\begin{align}
A = \frac{\sum_{j=1}^{m-1} p_j a_j^n}{p_m a_m^n}, \
B = \frac{\sum_{j=1}^{m-1} p_j a_j^{n-1}}{p_m a_m^{n-1}}, \
C = \frac{\sum_{j=1}^{m-1} p_j a_j^{n+1}}{p_m a_m^{n+1}}.
\end{align}
To proceed, we need the following result. The proof is given at the end.
Fact 1: Let $x, y, z \ge 0$ and $yz \ge x^2$. Then
$$\frac{(1+x)^2}{(1+y)(1+z)} \ge 1 - (y + z - 2x).$$
Let us proceed. By Cauchy-Bunyakovsky-Schwarz inequality, we have $BC \ge A^2$. By Fact 1, we have
$$1 - (B + C - 2A) \le f(n) \le 1$$
that is
$$1 - \frac{\sum_{j=1}^{m-1} p_j a_j^{n-1}(a_m - a_j)^2 }{p_m a_m^{n+1}}
\le f(n) \le 1.  \tag{1}$$
Remark: Actually, $1 - (B + C - 2A)$ is the Taylor expansion of $\frac{(1+A)^2}{(1+B)(1+C)}$ around $A=0, B=0, C=0$.
We have
\begin{align}
1 - \frac{\sum_{j=1}^{m-1} p_j a_j^{n-1}(a_m - a_j)^2 }{p_m a_m^{n+1}}
&\ge 1 - \frac{\sum_{j=1}^{m-1} p_j a_{m-1}^{n-1}(a_m - 0)^2 }{p_m a_m^{n+1}}\\
&= 1 - \frac{1-p_m}{p_m} \left(\frac{a_{m-1}}{a_m}\right)^{n-1}.
\end{align}
Thus, we have
$$1 - \frac{1-p_m}{p_m} \left(\frac{a_{m-1}}{a_m}\right)^{n-1} \le f(n) \le 1. \tag{2}$$
$\phantom{2}$
Proof of Fact 1: After clearing the denominators, it suffices to prove that
$$(y+z + yz - x)^2 - yz (1+y)(1+z) \ge 0.$$
From $yz \ge x^2$, we have $y + z \ge 2\sqrt{yz} \ge 2x$ and thus $0 \le y + z + yz - \sqrt{yz} \le y + z + yz - x$.
It suffices to prove that
$$(y+z + yz - \sqrt{yz})^2 - yz (1+y)(1+z) \ge 0$$
or
$$(y+z + yz)(\sqrt{y} - \sqrt{z})^2 \ge 0.$$
It is true. We are done.
