I need to approve or to disapprove the following statement :
if $(X_n)_{n \in \mathbb{N}}$ is a sequence of independent random variables and identically distributed, and $(u_n)_{n \in \mathbb{N}}$ is a sequence of real numbers such that $$\mathbb{P}(\limsup_n|\frac{1}{n}\sum_{k=1}^nX_k-u_n|<+\infty)>0$$
then $\mathbb{P}(\limsup_n\frac{1}{n}|X_{2n+1}-X_{2n}|<+\infty)=1$
I know that by kolmogorov $0$-$1$ law (since $(\frac{1}{n}|X_{2n+1}-X_{2n}|)_{n \in \mathbb{N}}$ is a sequence of independent random variable), we have $\mathbb{P}(\limsup_n\frac{1}{n}|X_{2n+1}-X_{2n}|<+\infty)=0 \ \ or \ \ 1,$
So which value does it take?
Suppose $P(\limsup_n \frac{1}{n}|X_{2n+1}-X_{2n}| < +\infty) < 1$. Then the $X_n$'s are not integrable. Let $\Delta_n = nu_n-(n-1)u_{n-1}$. Since $(\sum_{k=1}^n X_k -nu_n)-(\sum_{k=1}^{n-1} X_k-(n-1)u_{n-1}) = X_n-\Delta_n$, it suffices to show $\limsup_n \frac{|X_n-\Delta_n|}{n} = +\infty$ almost surely. So it suffices to show $\sum_{n=1}^\infty Pr(|X_n-\Delta_n| \ge nk) = +\infty$ for all $k \ge 1$. Clearly this is true if $|\Delta_n|/n \to +\infty$ along a subsequence, as there is some $M$ with $Pr(|X_1| \le M) > 0$. So we may assume $|\Delta_n| \le Cn$ for each $n$, and then we're done, since $\sum_{n=1}^\infty Pr(|X_n| \ge (C+k)n) = +\infty$ (since $E[|X_1|] = +\infty$; see proof of first part of Theorem 2.5.13 in Durrett's book).