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I want to prove the following theorem:

Suppose that we have given $s_{ij}\in[0,\infty]$ for each $i,j\in\mathbb{N}$. Define the sequence $(t_n)$ by: $t_1=s_{11}, t_2=s_{12}+s_{21},t_3=s_{13}+s_{22}+s_{31},\ldots$. Then: $$\sum_{n=1}^{\infty}{t_n}=\sum_{i=1}^{\infty}(\sum_{j=1}^{\infty}{s_{ij}})$$


Herefore some information: we suppose that $\sum_{n=1}^{\infty}{x_n}\in[0,\infty]$ thus the sum always exists.


HINT: Prove first that $$\sum_{n=1}^{N}{t_n}\leq\sum_{i=1}^{N}(\sum_{j=1}^{N}{s_{ij}})$$ and $$\sum_{n=1}^{N+M}{t_n}\geq\sum_{i=1}^{N}(\sum_{j=1}^{M}{s_{ij}})$$ then show: $$\sum_{j}(\sum_{i}{s_{ij}})=\sum_{i}(\sum_{j}{s_{ij}})$$


Question: Can someone help me with this theorem?! It seems to be very difficult and i don't know how to handle this with all these sums :(

Thank you!

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  • $\begingroup$ $$t_1 + t_2 + \ldots + t_N = s_{11}+(s_{12}+s_{21})+(s_{13}+s_{22}+s_{31})+\ldots+(s_{1N}+s_{2,N-1}+ \ldots +s_{N1}) = (s_{11}+ s_{12}+ \ldots +s_{1N}) + (s_{21}+ s_{22}+ \ldots +s_{2,N-1}) + \ldots + (s_{N-1,1}+s_{N-1,2})+s_{N1} \le (s_{11}+ s_{12}+ \ldots +s_{1N}) + (s_{21}+ s_{22}+ \ldots +s_{2N}) + \ldots + (s_{N1}+ s_{N2}+ \ldots +s_{NN}).$$ $\endgroup$
    – njguliyev
    Sep 8, 2013 at 15:19
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    $\begingroup$ Draw a picture of $\mathbb N\times\mathbb N$ and look at the indices each $t_N$ includes. These form a triangle $T_N$. Find two squares $S_N$ and $S'_N$ such that $S_N\subseteq T_N\subseteq S'_N$. This should give you an idea about the proof. $\endgroup$
    – Did
    Sep 8, 2013 at 15:22

1 Answer 1

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The hint should be considered in connection with the picture below:

$$\begin{array}{cc} \color{blue}{s_{11}}&\color{red}{s_{12}}&\color{green}{s_{13}}&\ldots&s_{1,n}&\ldots&s_{1,2n-3}&s_{1,2n-2}&\color{brown}{s_{1,2n-1}}\\ \color{red}{s_{21}}&\color{green}{s_{22}}&s_{23}&\ldots&s_{2,n}&\ldots&s_{2,2n-3}&\color{brown}{s_{2,2n-2}}&s_{2,2n-1}\\ \color{green}{s_{31}}&s_{32}&s_{33}&\ldots&s_{3,n}&\ldots&\color{brown}{s_{3,2n-3}}&s_{3,2n-2}&s_{3,2n-1}\\ \vdots&\vdots&\vdots&\ddots&\vdots&\ddots&\vdots&\vdots&\vdots\\ s_{n,1}&s_{n,2}&s_{n,3}&\ldots&\color{brown}{s_{n,n}}&\ldots&s_{n,2n-3}&s_{n,2n-2}&s_{n,2n-1}\\ \vdots&\vdots&\vdots&\ddots&\vdots&\ddots&\vdots&\vdots&\vdots\\ s_{2n-3,1}&s_{2n-3,2}&\color{brown}{s_{2n-3,3}}&\ldots&s_{2n-3,n}&\ldots&s_{2n-3,2n-3}&s_{2n-3,2n-2}&s_{2n-3,2n-1}\\ s_{2n-2,1}&\color{brown}{s_{2n-2,2}}&s_{2n-1,3}&\ldots&s_{2n-2,n}&\ldots&s_{2n-2,2n-3}&s_{2n-2,2n-2}&s_{2n-2,2n-1}\\ \color{brown}{s_{2n-1,1}}&s_{2n-1,2}&s_{2n-1,3}&\ldots&s_{2n-1,n}&\ldots&s_{2n-1,2n-3}&s_{2n-1,2n-2}&s_{2n-1,2n-1} \end{array}$$

Note that:

$$\begin{align*} &\color{blue}{t_1=s_{11}}\\ &\color{red}{t_2=s_{12}+s_{21}}\\ &\color{green}{t_3=s_{13}+s_{22}+s_{31}}\\ &\quad\,\,\vdots\\ &\color{brown}{t_n=s_{1n}+s_{2,n-1}+\ldots+s_{n-1,2}+s_{n1}} \end{align*}$$

In particular, compare the picture with the sums

$$\sum_{i=1}^n\sum_{j=1}^ns_{ij}\;,\quad\sum_{i=1}^nt_i\;,\quad\text{and}\quad\sum_{i=1}^{2n-1}\sum_{j=1}^{2n-1}s_{ij}\;.$$

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