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 $j\to\infty$.