Let $n \in \mathbb{N}$ be non-negative. Show that if $E[|X|^{n}]$ is finite, then:
$\lim\limits_{x\rightarrow\infty}x^{n}P(|X|\geq x) = 0$
Attempt at Solution
By Markov's inequality, we have:
$x^{n}P(|X|\geq x) \leq E[|X|^{n}]$
I'm not sure how to proceed from here. Is it true that: $\lim\limits_{x\rightarrow\infty}E[|X|^{n}] = 0$? If so, I can't understand why that is.
$E[|X|^n]$ does not depend on $x$. If it helps you, write Markov's inequality using $\epsilon$ instead of $x$ to avoid confusion:
$$P(|X|\geq \epsilon) \leq \frac{E[|X|^p]}{\epsilon^p}.$$
This is true for all $\epsilon>0$, and the numerator is independent of $\epsilon$.
To prove your result, though, you need more than just Markov. Here's a sketch:
$$E[|X|^n]=\int_0^\infty P(|X|^n>x)dx=\int_0^\infty nu^{n-1}P(|X|>u)du,$$
where $u=x^{1/n}$. Now you need $u^{n-1}P(|X|>u)=o(1/u)$, which will imply your limit (i.e. compare to $f(u)=1/u$ which is not integrable). There are some details here that need to be filled in.