Let $(Z_n)_{n\geq 1}$ be a sequence of iid random variables in $(\Omega,\mathcal{F},\mathbb{}P)$ such that $Z_n\sim N(0,1)$, what can be said about the following random series almost-surely convergence?
- $\sum_{k\geq1}\frac{Z_k}{k}$
- $\sum_{k\geq1}\frac{Z_k}{\sqrt{k}}$
- $\sum_{k\geq1}\frac{Z_k}{\ln(k)}$
It's easily seen that
$$ \mathbb{V}ar\left(\frac{Z_k}{k}\right)=\frac{1}{k^2} $$ and thus $$ \sum_{k\geq1}\mathbb{V}ar\left(\frac{Z_k}{k}\right)=\sum_{k\geq1}\frac{1}{k^2}<\infty $$ such that the convergence comes from the fact that the series is a $p$-series with $p>1$ and can be easily verified through the integral test for series. As the series of variances converge we may apply the Kolmogorov 1-series theorem and conclude that $$ \sum_{k\geq1}\frac{Z_k}{k} < \infty \quad\mathbb{P}-\text{a.s.} $$
As for the other two, the variance series clearly diverges because we get the harmonic series and a Bertrand series with $\alpha=0$ and $\beta=1$ and thus another way to verify if they diverge/converge is to check the Kolmgorov three-series theorem conditions. My attempt is as follows
\begin{align} \mathbb{E}\left[\frac{Z_k}{\sqrt{k}} I\left\{\left\lvert\frac{Z_k}{\sqrt{k}}\right\rvert\leq c\right\}\right]&=\mathbb{E}\left[\frac{Z_k}{\ln(k)}I\left\{\left\lvert\frac{Z_k}{\ln(k)}\right\rvert<c\right\}\right]=0 \\ \mathbb{V}ar\left[\frac{Z_k}{\sqrt{k}}I\left\{\left\lvert\frac{Z_k}{\sqrt{k}}\right\rvert\leq c\right\}\right]&=\frac{1}{n}\mathbb{E}[Z_k^2 I\left\{\left\lvert Z_k\right\rvert\leq \sqrt{k}c\right\}] \\ \mathbb{V}ar\left[\frac{Z_k}{\ln(k)}I\left\{\left\lvert\frac{Z_k}{\ln(k)}\right\rvert\leq c\right\}\right]&=\frac{1}{\ln(k)^2}\mathbb{E}[Z_k^2 I\left\{\left\lvert Z_k\right\rvert\leq \ln(k)c\right\}] \end{align} such that both variances are composed of a deterministic fraction and an expectation of a variable dominated by an integrable one, $Z_1$, thus we may apply the dominated convergence theorem to obtain that
\begin{align} \mathbb{E}[Z_k^2 I\left\{\left\lvert Z_k\right\rvert\leq \sqrt{k}c\right\}] &\to \mathbb{E}[Z_k^2] =1\\ \mathbb{E}[Z_k^2 I\left\{\left\lvert Z_k\right\rvert\leq \ln(k)c\right\}] &\to \mathbb{E}[Z_k^2] =1 \end{align} So we get the following two series of variances for the truncated variables and we may choose $\epsilon=1/2$ in the limits above in order to obtain that \begin{align} \sum_{k\geq 1}\frac{a_k}{k} &= \sum_{1\leq k < n_0}\frac{a_k}{k} + \sum_{k \geq n_0}\frac{a_k}{k} \\ &> \sum_{1\leq k < n_0}\frac{a_k}{k} + \frac{1}{2}\sum_{k \geq n_0}\frac{1}{k} = +\infty\\ \sum_{k\geq 1}\frac{a_k}{\ln(k)^2} &= \sum_{1\leq k < n_0}\frac{a_k}{\ln(k)^2} + \sum_{k \geq n_0}\frac{a_k}{\ln(k)^2} \\&> \sum_{1\leq k < n_0}\frac{a_k}{\ln(k)^2} + \frac{1}{2}\sum_{k \geq n_0}\frac{1}{\ln(k)^2} =+\infty \end{align} such that $(a_k)_{k\geq 1}$ is just short for the sequence of expectations and we obtain that the variance of the truncated variables diverge for every $c>0$ and by the contrapositive of the 3-series statemente we get that series (2) and (3) diverge almost-surely.
Is my attempt correct? Do the above statements generalize for something like the series $\frac{Z_k}{g(k)}$ always diverge for $g(k)=O(-)$? My guess is that if $g$ is $o(n)$ then the series diverges. I'm asking for a lot of solution-verifications lately as I'm going to have an exam soon and I'm trying to be really thorough with my solutions.