If $X_n \to 0$ in $L^p$ then $S_n/n \to 0$ in $L^p$ Let $X_n; n \geq 1$ be any sequence of random variables and $p \geq 1$. Denote $$S_n = \sum_{i=1}^n X_i.$$ Show that if $X_n \to 0$ in $L^p$ then $S_n/n \to 0$ in $L^p$. Also show that this is false if $p < 1$.
To start:
If $X_n \to 0$ in $L^p$ then by definition, $\lim_{n \to \infty} \mathbb{E}(|X_n|^p) = 0$. Then we have
\begin{align*}
\lim_{n \to \infty} \mathbb{E}\left(\left|\frac{1}{n} \sum_{i=1}^n X_i \right|^p\right) &= 
\lim_{n \to \infty} \frac{1}{n^p} \mathbb{E}\left(\left|\sum_{i=1}^n X_i \right|^p\right) \\
&\leq \lim_{n \to \infty} \frac{1}{n^p} \mathbb{E}\left[\left(\sum_{i=1}^n \left|X_i \right|\right)^p\right]
\end{align*}
I am unsure how to progress.
 A: For $p\ge 1$: Basic result in real analysis: If $a_n$ is a sequence of real numbers and $a_n\to L,$ then
$$\frac{a_1 + \cdots + a_n}{n} \to L.$$
Use this with the following:
$$\|\frac{S_n}{n}\|_p = \frac{1}{n }\|S_n\|_p \le \frac{1}{n}\sum_{k=1}^{n}\|X_k\|_p .$$
A: I will help you with the case $p>1$. Since $X_n\to 0$ in $L^p$, we have in hindsight that $E|X_n|^p<\infty$ for all $n\geq 1$. Then, from the $L^p$ convergence, it follows that for every $\epsilon>0$, there exists a natural number $N=N(\epsilon)$ such that
\begin{equation}
E|X_n|^p <\epsilon\quad \text{for all }n\geq N.
\end{equation}
Fixing $\epsilon=0.5$, it follows that there exists $N$ such that $E|X_n|^p<0.5$ for all $n\geq N$. Let $c=\max\{E|X_1|^p,\ldots,E|X_{N-1}|^p,0.5\}$. Then, we have $E|X_n|^p\leq c$ for all $n\geq 1$. In other words, the sequence $(E|X_n|^p)_{n\geq 1}$ is bounded.
Since $f(t)=t^p$ is convex for all $p\geq 1$ and $t\geq 0$, we have by Jensen's inequality and Minkowski's inequality that $$E\left(\left|\frac{1}{n} \sum_{i=1}^n X_i \right|^p\right)\leq \frac{1}{n^p}\sum\limits_{i=1}^nE|X_i|^p\leq \frac{c}{n^{p-1}}.$$
The above proves the desired $L^p$ convergence.
