Suppose $\{X_n\}$ a sequence of random variables.
If $\sum_{n=1}^\infty P(|X_n|>n)< \infty$
Prove that $$\limsup_{n\to\infty}\frac{ |X_n|}{n} \le1 $$ almost surely
What i have done so far:
I thought using the Borel-Cantelli lemma could lead me somewhere, but i didn't have any luck.
From Borel-Cantelli lemma we know that if $\sum_{n=1}^\infty P(|X_n|>n)< \infty$ then $P(|X_n|>n)=0$
How could I proceed?
I would appreciate any help, advice. Thank you all very much in advance for your time and concern.
Best Answer
The first Borel-Cantelli lemma yields $$\mathbb P\left(\limsup_{n\to\infty}|X_n|>n\right)=0. $$ As for each $n$ $$\{|X_n|>n\}\subset \bigcup_{k=n}^\infty \{|X_k|>k\}, $$ it follows that \begin{align} 0&=\mathbb P\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty \{|X_k|>k\}\right) \\&=\mathbb P\left(\lim_{n\to\infty}\bigcup_{k=n}^\infty \{|X_k|>k\} \right)\\ &=\lim_{n\to\infty}\mathbb P\left(\bigcup_{k=n}^\infty \{|X_k|>k\} \right)\\ &\geqslant\lim_{n\to\infty}\mathbb P(|X_n|>n) \end{align} and hence $\lim_{n\to\infty}\mathbb P(|X_n|>n)=0$. Further, $$\bigcap_{k=n}^\infty \{|X_k|>k\}\subset\{|X_n|>n\} $$ so that \begin{align} \mathbb P\left(\limsup_{n\to\infty} |X_n|\leqslant n\right) &= \mathbb P\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty\{|X_k|\leqslant k\}\right)\\ &=1 - \mathbb P\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty \{|X_k|>k\} \right)\\ &=1 - \mathbb P\left(\lim_{n\to\infty} \bigcap_{k=n}^\infty\{|X_k|>k\} \right)\\ &=1 - \lim_{n\to\infty}\mathbb P\left(\bigcap_{k=n}^\infty\{|X_k|>k\}\right)\\ &\geqslant1 - \lim_{n\to\infty}\mathbb P(|X_n|>n)\\ &= 1, \end{align} which implies that $$\mathbb P\left(\limsup_{n\to\infty} \frac{|X_n|}n\leqslant 1 \right)=1. $$