different type of convergence of random variable with summation

Question

I have the following problem:
Let $\xi_n: (\Omega, \mathcal F, \mathrm P) \to (\mathbb R^1, Bor)$
I need to find a connections between:
1. $\xi_n \to \xi$, such that $\sum_{n=1}^\infty|\xi_n - \xi| \lt \infty$ a.s.
2. $\xi_n \to \xi$, such that $\sum_{n=1}^\infty|F_{\xi_n}(x) - F_\xi(x)| \lt \infty$, $\forall x$ of continuity of distribution of $F_\xi$
What I try to do:
1. Prove that first convergence imply almost sure covergence (it quite simple: $\mathrm P(\omega \in \Omega: \sum_{n=1}^\infty|\xi_n - \xi| \lt \infty) = 1$, then I say, since sum converge absolutely, we have $|\xi_n - \xi| \to 0$, and this is the definition of almost sure converge.
2. Since $\xi_n$ converge to $\xi$ a.s., $\xi_n$ also converge in distribution, so $|F_{\xi_n}(x) - F_\xi(x)| \to 0$
3. What I can't proof, is speed of distributions converge.
Did I choose the right strategy to proof the statement?
To the opposite direction ($2 \implies 1$):
I can simply take a sequence from here: Convergence of random variables in probability but not almost surely. , it'll also converge in terms $\sum_{n=1}^\infty|F_{\xi_n}(x) - F_\xi(x)|$, but it won`t converge in a.s. sence and hence in terms $\sum_{n=1}^\infty|\xi_n - \xi|$

Answer 1

First condition doesn't imply second, we need some kind of uniform boundeness. For example, let $\xi$ and $\psi$ be uniform on $[0, 1]$ and independent, and $\xi_n = \begin{cases}\xi,\ \psi > \frac{1}{n}\\ 2,\ \psi \leq \frac{1}{n} \end{cases}$.

We have $\sum_n |\xi_n - \xi| < \frac{2}{\psi}$ (as there are as most $\frac{1}{\psi}$ non-zero terms, and each is at most $2$), so $\sum_n |\xi_n - \xi|$ converges a.s.

But $F_{\xi_n}(1) = 1 - \frac{1}{n}$, so $\sum_n|F_{\xi_n}(1) - F(1)| = \sum_n \frac{1}{n}$ diverges.