Let $A=(a\_{ij})$ be an infinite doubly stochastic matrix. Is the following statement true ?
$$
\lim\_{n\to\infty}\frac{1}{n}\sum\_{i=1}^n\sum\_{j=1}^na\_{ij} >0
$$


Any reference or comment on this is much appreciated.



A question on the partial sum of infinite doubly stochastic matrix
The answer is no. Indeed, let 
\begin{equation\*}
 s\_n:=\sum\_{i=1}^n\sum\_{j=1}^na\_{ij},\quad b\_n:=\frac{s\_n}{n}=\frac{1}{n}\sum\_{i=1}^n\sum\_{j=1}^na\_{ij}. 
\end{equation\*}
Then $0\le s\_n\le\sum\_{i=1}^n\sum\_{j=1}^\infty a\_{ij}=n$, and hence we always have 
\begin{equation\*}
 0\le\liminf\_{n\to\infty}b\_n\le\limsup\_{n\to\infty}b\_n\le1. 
\end{equation\*}


After these preliminaries, let us give an example with 
\begin{equation\*}
 \liminf\_{n\to\infty}b\_n=0,\quad \limsup\_{n\to\infty}b\_n=1. 
\end{equation\*}
Let $A=\text{diag}\,(A\_1,A\_2,\dots)$ be a block-diagonal matrix, where for each natural $k$ the $k$th diagonal block $A\_k$ is the $\ell\_k\times\ell\_k$ (permutation) matrix with $1$'s only on the secondary diagonal. So, letting $n\_k:=\sum\_{j=1}^k\ell\_j$ (with $n\_0=0$), we have $a\_{ij}=1$ if for some natural $k$ one has $n\_{k-1}+1\le i\le n\_k$ and $i+j=n\_{k-1}+1+n\_k[=2n\_{k-1}+1+\ell\_k]$, and $a\_{ij}=0$ otherwise. Then $A$ is doubly stochastic. 
Moreover, for each natural $k$, we have $b\_{n\_k}=1$ and hence indeed $\limsup\_{n\to\infty}b\_n=1$. On the other hand, assuming that the $\ell\_k$'s are even, for $m\_k:=n\_{k-1}+\ell\_k/2$ and $k\to\infty$ one has 
\begin{equation\*}
 b\_{m\_k}=\frac{n\_{k-1}}{m\_k}\to0 
\end{equation\*}
as $k\to\infty$ 
if e.g. $\ell\_k\sim k^k$ and hence $n\_k\sim k^k$ and $m\_k\sim k^k/2$; so, indeed here $\liminf\_{n\to\infty}b\_n=0$. 


Furthermore, since $0\le s\_{n+1}-s\_n\le2$, we always have the quasi-continuity condition $b\_{n+1}-b\_n\to0$ as $n\to\infty$. So, with a little extra effort, we can see that in the above example for each $c\in[0,1]$ there is an increasing sequence $(q\_j)$ of natural numbers such that $b\_{q\_j}\to c$ as $j\to\infty$.


