It's about probability and I was doing my homework. There are two questions.
1) $P(|X_{n}-\log n|=0\text{ i.o.}n)=1$. Is this equivalent to $\lim_{n\rightarrow\infty}\frac{X_{n}}{\log n}=1$ a.s.?
2) What is the definition of $\lim\sup_{n\rightarrow\infty}X_{n}=1$ a.s.?
Note that $X_{n}$ are random variables, a.s. means almost surely, and i.o is infinitely often. Thanks in advance.
$\{X_{n}=\log(n)\}$ infintely often means that given any $n\geq 1$, there exists a $k\geq n$ such that $X_{k}=\log(k)$ or $\frac{X_{k}}{\log(k)}=1$. Thus, \begin{equation} \sup\limits_{k\geq n}\frac{X_{k}}{\log(k)}\geq 1 \text{ a.s.}, \end{equation} and since this is true for every $n\geq 1$, we have \begin{equation} \inf\limits_{n\geq 1}\sup\limits_{k\geq n}\frac{X_{k}}{\log(k)}\geq 1\text{ a.s.}, \end{equation} which is another way of saying that $\limsup\limits_{n\rightarrow \infty}\frac{X_{n}}{\log(n)}\geq 1\text{ a.s}$.
In other words, the event $\{X_{n}=\log(n)\text{ i.o.}\}$ implies the event $\left\lbrace\limsup\limits_{n\rightarrow \infty}\frac{X_{n}}{\log(n)}\geq 1\right\rbrace$, and we therefore have \begin{equation} 1=P\left(\{X_{n}=\log(n)\text{ i.o.}\}\right)\leq P\left(\left\lbrace\limsup\limits_{n\rightarrow \infty}\frac{X_{n}}{\log(n)}\geq 1\right\rbrace\right), \end{equation} thus yielding $P\left(\left\lbrace\limsup\limits_{n\rightarrow \infty}\frac{X_{n}}{\log(n)}\geq 1\right\rbrace\right)=1$. However, what you have in the same question next is saying that the limit is $1$ with probability $1$. Clearly, there is a difference between saying that the limsup is $\geq 1$ with probability $1$ versus saying that the limit is $1$ with probability 1.
The question in (b) says that $P\left(\{\limsup\limits_{n\rightarrow \infty}X_{n}=1\}\right)=1$. You can use the earlier definition of limsup that I have written above to interpret further.