I want to prove the following:
Consider a sequence of $(Z_n)_{n \in \mathbb{N}}$ of independent random variables such that $Z_n \sim \mathcal{N}(0,\sigma_n^2)$. Now let $S_n = \sum_{j=1}^n Z_i$ and $\gamma_n^2 := \sum_{j=1}^n\sigma_j^2$ and assume that $\lim_{n \to \infty} \gamma_n^2 < \infty$. Then for any $p > 2$ $\lim_{n \to \infty} n^{-1/p}S_n=0$ a.s.
Now, my first observation was that $S_n \sim \mathcal{N}(0,\gamma_n^2)$ as a sum of independent random variables and thus $S_n' := n^{-1/p}S_n \sim \mathcal{N}(0,\frac{\gamma_n^2}{n^{2/p}})$. Hence for any $\varepsilon > 0$ by the Markov inequality we get for $q \in \mathbb{N}$ with $q \geq p$
$\mathbb{P}(\vert S_n' \vert > \varepsilon) \leq \frac{\mathbb{E}(\vert S_n'\vert^q)}{\varepsilon^q} \leq \frac{\mu_q}{\varepsilon^q} \frac{(\gamma_n^{2})^q}{n^{2q/p}} \leq \frac{\mu_q}{\varepsilon^q} \frac{(\gamma_n^{2})^q}{n^{2}} $
where $\mu_q$ is the q-th moment of the standard Gaussian, see here. Thus
$\sum_{n=1}^\infty \mathbb{P}(\vert S_n' \vert > \varepsilon) \leq \frac{\mu_q}{\varepsilon^q} \sum_{n=1}^\infty\frac{(\gamma_n^{2})^q}{n^{2}} < \infty$
since $ \gamma_n^2 < c$ for sufficient large $n$, which implies $\lim_{n \to \infty} S_n' = 0$ a.s. by the Borel-Cantelli lemma.
First question: Does this look correct? And second: Is there a more elegant way to show that?
Best Answer
Actually, $S'_n$ has the same distribution as $ \gamma_n n^{-1/p}N$, where $N$ has a standard normal distribution hence $$ \mathbb E\left[\left\vert S'_n\right\rvert^q\right]=\left(\gamma_n n^{-1/p}\right)^q\mu_q\leqslant \left(\sup_{\ell\geqslant 1}\gamma_k\right)^q \mu_qn^{-q/p}$$ hence you have to choose $q>p$.
Actually, it is possible to show that $\left(S_n\right)_{n\geqslant 1}$ converges almost surely (by this result, it suffices to show the convergence in probability, which can be done by showing the convergence in $\mathbb L^2$ for instance).
Therefore,$S_n/n^{\beta}\to 0$ almost surely for each positive $\beta$.