Question about sums and double sums 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!
 A: 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
$$\sum_{i=1}^n\sum_{j=1}^ns_{ij}\;,\quad\sum_{i=1}^nt_i\;,\quad\text{and}\quad\sum_{i=1}^{2n-1}\sum_{j=1}^{2n-1}s_{ij}\;.$$
