Almost sure convergence of subsequence If we have for a sequence of identically independent variables with  $\limsup \{|X_n|>n\}$ happens almost surely. Can we conclude that $\limsup \{n^{-1}|\sum X_j|>1\}$ happens a.s.? One of the problem I was working on seem to need this to conclude but I don't see why this is true. Any help?
 A: Yes, it is true.  Here is a proof of a generalization:  
Claim:
Suppose $\{X_n\}_{n=1}^{\infty}$ are independent and identically distributed (i.i.d.) and there is an $\alpha>0$ such that 
$$ P[\{|X_n|> \alpha n\} \quad i.o.] = 1$$
where "i.o." stands for "infinitely often."  Then for any $\beta>0$ we have: 
$$P[\{|X_n|>\beta n\} \quad i.o.]=1 \quad (Eq. *)$$
and 
$$P\left[\left\{\left|\frac{1}{n}\sum_{i=1}^n X_i\right|>\beta \right\} \quad i.o.\right] = 1 \quad (Eq. **)$$

Fact 1:
Let $Y$ be a nonnegative random variable. Fix $\theta>0$. Then $E[Y]=\infty$ if and only if 
$$ \sum_{n=1}^{\infty}P[Y>\theta n] = \infty$$
Proof: This can be proven by $E[Y]=\int_0^{\infty} P[Y>t]dt$ and bounding the integral by a sum. 

Proof of Claim:
Suppose $\{X_n\}_{n=1}^{\infty}$ are i.i.d. and there is an $\alpha>0$ that satisfies $P[\{|X_n|> \alpha n\} \quad i.o.]=1$. 
By Borel-Cantelli we know
$$ \sum_{n=1}^{\infty} P[|X_n|> \alpha n] = \infty$$
since if the sum were finite, $\{|X_n|> \alpha n\}$ would happen only finitely often (with prob 1).  Since $\{X_n\}$ are i.i.d. we know $P[|X_n|> \alpha n]=P[|X_1|> \alpha n]$ and so 
$$ \sum_{n=1}^{\infty} P[|X_1|> \alpha n] = \infty$$ 
It follows from Fact 1 that $E[|X_1|]=\infty$ and so for any $\theta>0$
$$ \sum_{n=1}^{\infty} P[|X_1|>\theta n] = \infty$$
and so 
$$ \sum_{n=1}^{\infty} P[|X_n|>\theta n] = \infty$$
Since $\{X_n\}$ are mutually independent, by Borel-Cantelli this implies 
$$ P[\{|X_n|>\theta n\} \quad i.o.] = 1 \quad (Eq. ***)$$
This proves (Eq. *). 
Now fix $\beta>0$.  Define $L_n = \frac{1}{n}\sum_{i=1}^n X_i$.  We have by the Kolmogorov 0-1 law that 
$$ P[\{|L_n|>\beta \} \quad i.o.]  \in \{0,1\}$$
If the probability is 1 then we are done. Now suppose the probability is zero (we reach a contradiction).  Then $|L_n|\leq \beta$ for all but finitely many indices $n$ (with prob 1). But if we define $\theta = 3\beta$ and use (Eq. ***), we know $\{|X_n|>3\beta n\}$ happens infinitely often (with prob 1).  Thus, with prob 1, the events $\{|X_{n+1}|>3\beta (n+1) \}\cap\{|L_n|\leq \beta \}$ occur infinitely often.  Any such time this event occurs it must be that $|L_{n+1}|>\beta$ because: 
$$ |L_{n+1}| = \left|\frac{nL_n}{n+1} + \frac{X_n}{n+1}\right| \geq \frac{|X_n|-|nL_n|}{n+1}> \frac{3\beta n - \beta n}{n+1} \geq \beta $$
Thus
$$P[|L_{n+1}|> \beta \quad i.o. ] = 1$$
which contradicts our assumption that the probability is 0. This proves (Eq. **). $\Box$
