# Showing that $\lim_{l\to\infty}\frac{E\left(X^{l+1}\right)}{E\left(X^l\right)}=n$

Let $$X$$ be a random variable that has the discrete uniform distribution on $$\{1,2,...,n\} (n\geq 2)$$, i.e., $$P(X=k)=\frac{1}{n}$$

For $$k\in\{1,2,...,n\}$$

I would like to prove that $$\lim_{l\to\infty}\frac{E\left(X^{l+1}\right)}{E\left(X^l\right)}=n$$

I know that $$\frac{E\left(X^{l+1}\right)}{E\left(X^l\right)}=\frac{\sum^n_{k=1}k^{l+1}\frac{1}{n}}{\sum^n_{k=1}k^l\frac{1}{n}}=\frac{\sum^n_{k=1}k^{l+1}}{\sum^n_{k=1}k^l}$$

And now I would like to simplify it so that the terms which depend on $$l$$ converge to $$0$$ as $$l\to\infty$$, however I am not sure how to do this.

• Your problem is a manifestation of a more general phenomena: If $(\Omega,\mathscr{F},\mathbb{P})$ is a probability space and $X$ is a random variable. Then $$\lim_{p\rightarrow\infty}\frac{\|X\|^{p+1}_{p+1}}{\|X\|^p_p}=\|X\|_\infty$$ Aug 24 at 15:20

We have that as $$l\to \infty$$
$$\frac{\sum^n_{k=1}k^{l+1}}{\sum^n_{k=1}k^l}=\frac{n^{l+1}}{n^l}\frac{\sum^n_{k=1}\left(\frac{k}n\right)^{l+1}}{\sum^n_{k=1}\left(\frac{k}n\right)^l} \to n \cdot \frac 1 1=n$$
$$\frac{\sum^n_{k=1}k^{l+1}}{\sum^n_{k=1}k^l}\sim \frac{n^{l+1}}{n^l}=n$$