Let $N$ denote the set of natural numbers. Let $(x_{n,m})_{n,m \in N}$ be a double sequence of positive numbers that satisfies:
1) $x_{n+1,m} \leq x_{n,m}$, for every $m,n \in N$.
2) $\sum_{n=1}^{\infty} x_{n,m} = 1$, for every $m \in N$.
3) For every $n \in N$, $\lim_{m \to \infty} x_{n,m} = x_n$ for some positive number $x_n$.
Is it posible to show $\sum_{n=1}^{\infty} x_{n} = 1$?, If it is how can it be proved? If it is not, can you give a counterexample?
Note: I can’t find a way to use monotone, dominated or Vitali’s convergence theorems. The classic counterexample, $x_{n,m} = 1$ if $n=m$ and $x_{n,m} = 0$ if $n\neq m$ fails here due to hypothesis 1).
Thank you so much in advance
The answer is no. Consider the sequence $x_{n,m}$ given by $$x_{n,m}=\begin{cases}\frac{1}{m}\quad\text{ if }n\leq m\\0\quad\text{ else}\end{cases}$$ Then $x_{n+1,m}\leq x_{n,m}$ for all $n,m\in\mathbb{N}$, $\sum^{\infty}_{n=1}x_{n,m}=\sum^{m}_{n=1}\frac{1}{m}=1$ for all $m\in\mathbb{N}$ and $\lim_{m\rightarrow\infty}x_{n,m}=0$.
If all $x_{n,m}$ and all $x_{n}$ have to be strictly positive the answer is still no. Consider the sequence $x_{n,m}$ given by $$x_{n,m}=\begin{cases}2^{-n-1}&\text{ if }n< m\\2^{-m-1}&\text{ if }m\leq n\leq m+2^{m}\\2^{-n+2^{m}-1}&\text{ if }n> m+2^{m}\end{cases}.$$ Clearly this sequence is strictly positive and satisfies condition $1$. For condition $2$ note that for all $m$ it holds that $$\sum_{n=1}^{\infty}x_{n,m}=\sum^{m}_{n=1}2^{-n-1}+\sum^{m+2^{m}}_{n=m+1}2^{-m-1}+\sum^{\infty}_{n=m+2^{m}+1}2^{-n+2^{m}-1}$$$$=\sum^{m}_{n=1}2^{-n-1}+\frac{1}{2}+\sum^{\infty}_{n=m+1}2^{-n-1}=\frac{1}{2}+\sum^{\infty}_{n=1}2^{-n-1}=1.$$ Finally note that $x_{n}=\lim_{m\rightarrow\infty}x_{n,m}=2^{-n-1}>0$ and we find that $$\sum^{\infty}_{n=1}x_{n}=\sum^{\infty}_{n=1}2^{-n-1}=\frac{1}{2}\neq1.$$