From the standard literature it is well known that for sequences of random variables $X\_{1, n} \stackrel{P}{\rightarrow} X\_1$ and $X\_{2, n} \stackrel{P}{\rightarrow} X\_2$ as $n \rightarrow \infty$ it holds that $(X\_{1, n}, X\_{2, n}) \stackrel{P}{\rightarrow} (X\_1, X\_2)$ for $n \rightarrow \infty$. Using a continuous mapping theorem argument this can be used to establish that $X\_{1, n} + X\_{2, n} \stackrel{P}{\rightarrow} X\_1 + X\_2$ for $n \rightarrow \infty$.


Question in general case
------------------------


Under which conditions can the arguments of the standard literature be extended from a simple sum of two sequences to a sum of sequences whose number of summands also increases to infinity as $n$ goes to infinity. Specifically, given a sequences $X\_{l, n}$, $l \in \mathbb{N}$, with $X\_{l, n} \stackrel{P}{\rightarrow} X\_l$ $(n \rightarrow \infty)$ for all $l\in \mathbb{N}$ under which conditions does it holds that
$$
\lim\_{n \rightarrow \infty} \sum\_{l = 1}^n X\_{l, n} \stackrel{P}{\rightarrow} \sum\_{l = 1}^\infty X\_{l}
$$
as $n \rightarrow \infty$?


Question in special case
------------------------


In my particular use case something more specific would also suffice. Though, I think the more general question is also interesting. If we know that
$X\_{l, n}$, $l \in \mathbb{N}$, are nonnegative random sequences, i.e. each sequence element is a nonnegative random variable, and $X\_{l, n} \stackrel{P}{\rightarrow} 0$ $(n \rightarrow \infty)$ for all $l \in \mathbb{N}$ under which conditions does it holds that
$$
\lim\_{n \rightarrow \infty} \sum\_{l = 1}^n X\_{l, n} \stackrel{P}{\rightarrow} 0
$$
as $n \rightarrow \infty$?
Possibly, what would also be interesting and might be easier to answer is when the sample mean converges to zero, i.e.,
$$
\lim\_{n \rightarrow \infty} \frac{1}{n} \sum\_{l = 1}^n X\_{l, n} \stackrel{P}{\rightarrow} 0
$$
as $n \rightarrow \infty$



Convergence in probability of series of random variables
$\newcommand\ep\varepsilon\newcommand\de\delta\newcommand{\P}[1]{\overset P{\underset{#1}\longrightarrow}}$What you need is the uniform summability (in probability).


Here are details: Let $Y\_{l,n}:=X\_{l,n}-X\_l$, so that $$Y\_{l,n}\P{n\to\infty}0 \tag{0}$$
for each $l$. We want to have
$$S\_{n,n}\P{n\to\infty}0,\tag{1}$$
where
$$S\_{m,n}:=\sum\_{l=1}^m Y\_{l,n}.$$
The mentioned sufficient uniform summability condition is that
$$S\_{n,n}-S\_{m,n}\P{n\ge m\to\infty}0. \tag{2}$$


Indeed, take any real $\de>0$ and $\ep>0$. Then, by (2), for some natural $m\_1$ we have the following implication:
$$n\ge m\ge m\_1\implies P(|S\_{n,n}-S\_{m,n}|>\ep/2)\le\de/2. \tag{3}$$
Also, (0) implies $S\_{m\_1,n}\P{n\to\infty}0$, so that for some natural $m\_2$ we have the following implication:
$$n\ge m\_2\implies P(|S\_{m\_1,n}|>\ep/2)\le\de/2. \tag{4}$$
Letting now $m\_3:=\max(m\_1,m\_2)$, by (3) and (4) we have
$$n\ge m\_3\implies 
P(|S\_{n,n}|>\ep)\le
P(|S\_{n,n}-S\_{m\_1,n}|>\ep/2)+P(|S\_{m\_1,n}|>\ep/2)\le\de. $$
So, (1) holds.


