Convergence of a random sequence without second moment

Question

Let $\{ \xi_n\}_{n \in \mathbb{N}}$ be a family of i.i.d. random variables whit finite mean, $\mathbb{E}\xi_1 = \mu < \infty$ and $\zeta > 2$.

How to prove that, almost sure, $$\lim_{n \to \infty} \frac{1}{n^{\zeta}}\sum_{i=1}^{n}(\xi_i -\mu)i^{\zeta-1} = 0~?$$

Note that if $\mathbb{E}\xi_1^2 < \infty$, we cam apply Kolmogorov's Maximal Inequality to get the result. But how proceed only with first moment? Looks like that we need a kind of truncation...

Thanks for you help.

Answer 1

Assume with loss of generality that $\mu=0$. Let $S_i=\sum_{k=1}^n\xi_k$ for $i\geqslant 1$ and $S_0=0$. Then $$ \sum_{i=1}^n\xi_i i^{\zeta-1}=\sum_{i=1}^n\left(S_i-S_{i-1}\right) i^{\zeta-1} =\sum_{\ell=1}^nS_\ell \ell^{\zeta-1}-\sum_{\ell=1}^{n-1}S_\ell\left(\ell+1\right)^{\zeta-1}= S_n n^{\zeta-1}+\sum_{\ell=1}^{n-1}S_\ell\left(\ell^{\zeta-1}-\left(\ell+1\right)^{\zeta-1}\right). $$ Then use the fact that $\left(S_\ell/\ell\right)_{\ell\geqslant 1}$ converges almost surely to $0$ to conclude.