I am struggling dealing with $o$ notation. Here $k, n$ are all natural numbers, why do we have $o\Big(\sum_{k=0}^{n-1}\frac{(k\log\log k)^{1/2}}{k+1}\Big)=o((n\log\log n)^{1/2})$ as $n\rightarrow\infty$? And another one with convergence, why does $\sum_{k=0}^{n-1}\frac{|X_k|}{k^2}$ converges in $L^2$ and a.s. as $n\rightarrow\infty$ ? Where $(X_k)_{k\in\mathbb{N}}$ are normally distributed random variables with mean $0$, variance $k$ and they have independent and stationary increment (but they themselves are not independent with each other).
Comment: I apologise that I changed the second part of the question a bit, since the original question was not formulated in the correct way.
I think I come up with an solution, it is much appreciated if you can help me double check that this is correct.
For the first small $o$ notation: since $k<n$, we have $\sum_{k=0}^{n-1}\frac{(k\log\log k)^{1/2}}{k+1}\leq \sqrt{\log\log n} \sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}\leq \sqrt{n\log\log n}$, thus the first result.
Regarding the second question: we use convergence of series of independent centered $L^2$ random variables. Here we use exchange the order of summation below $\sum_{k=1}^{n-1}\frac{|X_k|}{k^2}\leq \sum_{k=1}^{n-1}\sum_{i=1}^{k-1}\frac{|X_{i+1}-X_{i}|}{k^2}=\sum_{i=1}^{n-1}\sum_{k=i}^{n-1}\frac{|X_{i+1}-X_{i}|}{k^2}=\sum_{i=1}^{n-1}(\frac{1}{i}-\frac{1}{n-1})|X_{i+1}-X_{i}|$
Here $(X_{i+1}-X_{i})_{i\geq 1}$ is a sequence of independent random variables, with mean $0$ and variance $1$ (because of the independent and stationary increment). We only need to check if $\lim_{n\rightarrow\infty}\sum_{i=1}^{n-1}\mathbb{E}((\frac{1}{i}-\frac{1}{n-1})(X_{i+1}-X_{i}))^2$ converges, then we would have the result. And it is not hard to find that the above series $\leq \lim_{n\rightarrow\infty}(\frac{1}{n-1}+\frac{\log(n-1)}{n-1}+\frac{1}{n-1})=0$, hence the desired convergence. Thus we have $\sum_{k=1}^{n-1}\frac{X_k}{k^2}$ converging both in $L^2$ and a.s.