Stochastic convergence of a countable sum of stochastically convergent random variables Suppose we have countably many sequences $(X^i_n)_{n \in \mathbb{N}}$, $ i \in \mathbb{N}$, of random variables, such that every sequence $(X^i_n)$ converges in probability to a random variable $Y^i$. Is it then true that $\sum \limits_{i=1}^\infty X^i_n$ converges in probability to $\sum \limits_{i=1}^\infty Y^i$ as $ n \to \infty$ ? I assume that all random variables are non-negative, so both limits should be measurable and exist.
I tried to prove it analogously to the common proof for the sum of two stochastically convergent random variables, but i am not so sure if the argumentation is applicable.
I used $\epsilon \sum \limits_{i=1}^\infty \frac{1}{2^i} = \epsilon$ to conclude  $$\mathbb{P}\left(|\sum \limits_{i=1}^\infty X^i_n - \sum \limits_{i=1}^\infty Y^i|> \epsilon\right) \leq \mathbb{P}\left(\sum \limits_{i=1}^\infty |X^i_n - Y^i|> \epsilon\right) \leq \sum \limits_{i=1}^\infty \mathbb{P}\left(|X^i_n - Y^i|> \frac{\epsilon}{2^i}\right).$$
Now the summands in the latter all converge to zero by assumption. Does this mean that the whole term goes to zero for $n \to \infty$ ? And also i wonder if the last inequality is actually correct, since it first seemed valid to me that
$$\left\{\sum \limits_{i=1}^\infty |X^i_n - Y^i|> \epsilon\right\} \subset \cup_{i=1}^\infty \left\{ |X^i_n - Y^i|> \frac{\epsilon}{2^i} \right\} .$$
But is that really correct for an infinite series?
If not, is there a way to prove the claim or am I missing some well known theorem about that?
Thanks in advance!
 A: In general the claim does not hold. Let $(Y_n)_{n\geqslant 1}$ be a sequence which converges to $0$ in probability but such that $(nY_n)_{n\geqslant 1}$ does not converge in probability to $0$ (for example, $Y_n=\sum_{i=1}^nZ_i/n$, where $(Z_i)_{i\geqslant 1}$ is i.i.d., $Y_1$ is integrable and centered). Then let $X_n^i=Y_n\mathbf{1}_{n+1\leqslant i\leqslant 2n}$.
In this way, the inequality $\left\lvert X_n^i\right\rvert\leqslant \left\lvert Y_n\right\rvert$ guarantees that $X_n^i\to Y_i:=0$ in probability for each $i$ and $\sum_{i=1}^\infty X_n^i=\sum_{i=n+1}^{2n}Y_n=nY_n$ which does not converge to $0=\sum_{i=1}^\infty Y_i$ in probability.
Notice that in your approach, the series $\sum \limits_{i=1}^\infty \mathbb{P}(|X^i_n - Y^i|> \frac{\epsilon}{2^i})$ may fail to converge because $ \frac{\epsilon}{2^i}$ goes to $0$ and you cannot exploit the convergence in probability. In my example, this series would be
$\sum_{i=n+1}^{2n}\mathbb P(\lvert Y_n\rvert>\varepsilon 2^{-i})$ which  is bigger than $n\mathbb P(\lvert Y_n\rvert>\varepsilon 2^{-n})$ and this does not converge to $0$.
