For each $n\in\mathbb{N}$, consider real $n$ numbers $x_{i,n}$, $1\le i \le n$, given by $$\sum_{i=1}^{n}x_{i,n}=0,\\-1/n \le x_{i,n} \le 1-1/n.$$
I am trying to find value of $$\lim_{n\to \infty} \sum_{l=1}^{\lfloor n/p \rfloor}\textbf{1}_{lp\le n}x_{lp,n},$$
where $p$ denotes a prime and $\textbf{1}_{\lbrace\cdot \rbrace}$ denotes an indicator function.
Approach: Since $n$ is going to $\infty$, we can say $$\textbf{1}_{lp\le n}\equiv1$$ in the limit.
This leads to $$\lim_{n\to \infty} \sum_{l=1}^{\lfloor n/p \rfloor}x_{lp,n}=\sum_{i=1}^{\infty}x_{i,\infty} - x_{1,\infty}=-x_{1,\infty}.$$
Does this suggest the limit does not exist?
Let $n$ and $l\le\lfloor n/p\rfloor$ be any natural numbers. Then $lp\le n$, so $\textbf{1}_{lp\le n}=1$. Put $x_n=\sum_{l=1}^{\lfloor n/p \rfloor}\textbf{1}_{lp\le n}x_{lp,n}=\sum_{l=1}^{\lfloor n/p \rfloor} x_{lp,n}$. The (existence of the) limit $\lim_{n\to\infty} x_n$ depends on the numbers $x_i,n$ for $n\in\mathbb N$ and $1\le i\le n$. For instance, if all these numbers are zeroes then the limit is zero too. On the other hand, if $p=2$, and for each even $n$ and each natural $i\le n/2$ we have $x_{i,n}=\frac {(-1)^{i+n/2}}{n}$, then $x_n=\frac {(-1)^{n/2}}2$, so the limit does not exist.