Proving $\lim_{n\to\infty} \frac{EX^{n+1}}{EX^n}=\max\{x_1,\ldots,x_k\}$

Question

How to prove that $$\lim_{n\to\infty} \frac{E(X^{n+1})}{EX^n}=\max\{x_1,\ldots,x_k\}$$ for a discrete random variable $X\geq0$ taking values $x_1,\ldots,x_k$.

Answer 1

Suppose $x_1<\ldots<x_k$ and let $p_1,\ldots,p_k>0$ be the respective probabilities. Note \begin{equation} \frac{\mathbb{E}[X^{n+1}]}{\mathbb{E}[X^n]}=\frac{\sum_{i=1}^k p_ix_i^{n+1}}{\sum_{i=1}^k p_ix_i^{n}}= \frac{\sum_{i=1}^k p_ix_i(x_i/x_k)^n}{\sum_{i=1}^k p_i(x_i/x_k)^{n}}. \end{equation} Show that all terms but the $k$th tend to $0$ as $n\to \infty$, while the $k$th terms tend to $p_ix_i$ and $p_i$ respectively.