Does $\lim_{n \to \infty} E(X_n)/n = \lim X_n/n$? Let $\{X_n\}_{n=1}^{\infty}$ be a sequence of random variables (not i.i.d.) with $X_n > 0 $ and $E(X_n) < \infty$ for all $n$. Suppose that $\lim\limits_{n \to \infty} \frac{E(X_n)}{n} = L$, where $L \in \mathbb{R}$. Does it follow that
$$P \bigg( \lim_{n \to \infty} \frac{X_n}{n}  = L\bigg)  = 1 \;?$$
If not, I would be interested in a counterexample. Any help would be greatly appreciated.
 A: Note, that $E[X_n]$ is a number, while $X_n$ is a random variable.
Consider $X_n$ with $P(X_n=\pm n)=\frac{1}{2}$.
Then $E[X_n]=0$ for every $n$, but
$\frac{X_n}{n}$ is a random variable with
$P(\frac{X_n}{n}=\pm 1)=\frac{1}{2}$
This implies, that $\frac{X_n}{n}\rightarrow X$ with $P(X=\pm 1)=\frac{1}{2}$
Edit: For the case with $X_n>0$, the gamma distribution https://en.wikipedia.org/wiki/Gamma_distribution with parameter $\alpha=\beta=\frac{1}{n^2}$ might work, since $X_n$ has variance $n^2$, so $\frac{X_n}{n}$ has variance 1. Also, $E[X_n]=1$ for every $n$
A: I'm assuming that you're overloading notation by saying that $\lim_{n\to\infty}\frac{E_n}{n}$ is both a real number and a random variable which is constant and takes that same number as its only value.
In that case, no, it does not. You can use the central limit theorem to construct a counterexample: Let $Y_n$ be i.i.d. real valued random variables with finite mean $\mu$ and finite Variance $\sigma^2>0$. Define $X_n:=\sqrt{\frac{n}{\sigma^2}}\sum_{k=1}^n(Y_k-\mu)$. Then according to the central limit theorem, $\frac{X_n}{n}\to X$ in distribution, where $X\sim\mathcal N(0,1)$ (so especially $X\neq0$). But $E[X_n]=0$, so $\frac{E[X_n]}{n}\to0$.
