Let $(X_n)_n$ be a sequence of independent and identically distributed random variable, such that there exists $r>0$ such that $E[|X_1|^r]=+\infty.$ Let $(x_n)_n$ be a sequence of real numbers, $Y_n=\frac{1}{n^{1/r}}\sum_{k=1}^nX_k-x_n.$
1) Prove that
$$\limsup_n|Y_n|=+\infty \ a.s$$
2) Prove or disprove the following statement : $P(\left\{\limsup_nY_n=+\infty \right\}\cup\left\{\liminf_nY_n=-\infty \right\})=1.$
Attempt: I will use the fact that $(X_{2n},X_{2n+1})_n$ is a sequence of i.i.d random variables such that $E[|X_3-X_2|^r]=+\infty$ (since $E[|X_2|]=+\infty$).
Let $U_n=\dfrac{1}{n^{1/r}}\sum_{k=1}^nX_{2k}-x_n,W_n=\dfrac{1}{n^{1/r}}\sum_{k=1}^nX_{2k+1}-x_n.$
$\forall n,p \in \mathbb{N^*},\frac{1}{(p+n)^{1/r}}(X_{2(p+n)+1}-X_{2(p+n)})$ $=W_{p+n}+x_{p+n}-(\dfrac{p+n-1}{p+n})^{1/r}(W_{p+n-1}+x_{p+n-1})-U_{p+n}-x_{p+n}+(\dfrac{p+n-1}{p+n})^{1/r}(U_{p+n-1}+x_{p+n-1})$ $=W_{p+n}-(\dfrac{p+n-1}{p+n})^{1/r}W_{p+n-1}-U_{p+n}+(\dfrac{p+n-1}{p+n})^{1/r}U_{p+n-1}.$
We conclude that, $$\forall k,n,q \in \mathbb{N^*}P(\max_{1\leq p \leq q}\frac{1}{(p+n)^{1/r}}|X_{2(p+n)+1}-X_{2(p+n)}|>4k)\leq P(\max_{1 \leq p \leq q}|W_{p+n}|>k)+P(\max_{1\leq p\leq q}|W_{p+n-1}|>k)+P(\max_{1 \leq p \leq q}|U_{p+n}|>k)+P(\max_{1 \leq p \leq q}|U_{p+n-1}|>k),$$
and since $P(\max_{1 \leq p \leq q}|W_{p+n}|>k)=P(\max_{1 \leq p \leq q}|Y_{p+n}|>k)=P(\max_{1 \leq p \leq q}|U_{p+n}|>k),$
$P(\max_{1 \leq p \leq q}|W_{p+n-1}|>k)=P(\max_{1 \leq p \leq q}|Y_{p+n-1}|>k)=P(\max_{1 \leq p \leq q}|U_{p+n-1}|>k),$
meaning that $$\forall k,n,q \in \mathbb{N^*}P(\max_{1\leq p \leq q}\frac{1}{(p+n)^{1/r}}|X_{2(p+n)+1}-X_{2(p+n)}|>4k)\leq 2P(\max_{1 \leq p \leq q}|Y_{p+n}|>k)+2P(\max_{1\leq p\leq q}|Y_{p+n-1}|>k)$$ Taking the limit $q \to +\infty $
$$\forall k,n\in \mathbb{N^*}P(\sup_{p \in \mathbb{N}}\frac{1}{(p+n)^{1/r}}|X_{2(p+n)+1}-X_{2(p+n)}|>4k)\leq 2P(\sup_{p \in \mathbb{N}}|Y_{p+n}|>k)+2P(\sup_{p \in \mathbb{N}}|Y_{p+n-1}|>k)$$
Then taking the limit $n \to \infty$
$$\forall k\in \mathbb{N^*},1=P(\limsup_n\left\{\frac{1}{n^{1/r}}|X_{2n+1}-X_{2n)}|>4k\right\})\leq 4P(\limsup_n\left\{|Y_{n}|>k\right\})$$ Using Kolmogorov 0-1 law, $$\forall k\in \mathbb{N},P(\limsup_n\left\{|Y_{n}|>k\right\})=1$$ So $$\limsup_n|Y_n|=+\infty \ a.s$$
I am stuck on 2). Any ideas?