<s>[INST] How to sum an infinite matrix? Given this matrix that stretches to infinity to the right and up:
$$
        \begin{matrix}
        ...&...&...\\
        \frac{1}{4}& \frac{1}{8}& \frac{1}{16}&... \\
        \frac{1}{2} & \frac{1}{4}& \frac{1}{8}&... \\
        1 & \frac{1}{2}& \frac{1}{4}&... \\
        \end{matrix}
$$
I was trying to find the total sum of this matrix. I know the answer should be $4$. I came up with a different solution and a different answer. What is wrong with that solution? Here it is:
The first row sums to $2$. The second row to $2-1$. The third row to $2-1-\frac{1}{2}$ etc... So we get:
$$
        \begin{matrix}
        2&-1&-\frac{1}{2}&-\frac{1}{4}&-\frac{1}{8}&-\frac{1}{16}\\
        2&-1&-\frac{1}{2}&-\frac{1}{4}&-\frac{1}{8}\\
        2&-1&-\frac{1}{2}&-\frac{1}{4}\\      
        2&-1&-\frac{1}{2}\\
        2&-1 \\
        2 \\
        \end{matrix}
$$
Now for each "$2$" there is a diagonal that gives the sequence $2-1-\frac{1}{2}-\frac{1}{4}...=0$ (since the matrix goes on forever) Therefore, the sum of the matrix must be $0$!
Apparently that's wrong; but why? Thanks!
EDIT: I am looking for an answer to the question what is fundamentally wrong with my method plus an explanation for why that is wrong. I am not looking for an explanation of the correct method.
 [/INST] The ''triangular array'' is not  really an array but a column of values:
$ 2  $
$2-1=1$
$2-1-\frac{1}{2}=\frac{1}{2}$
$2-1-\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$
$2-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}=\frac{1}{8}$
$\cdots$
so there is not a diagonal
and the sum of these values is clearly $=4$

In other words, this is not a matrix, but  the ''infinite sum'' 
$$
2+\left(2-1\right)+\left(2-1-\frac{1}{2}\right)+\left(2-1-\frac{1}{2}-\frac{1}{4}\right)+ \cdots +\left(2-\sum_{i=1}^n\frac{1}{i}\right)+ \cdots
$$
and we cannot rearrange  or associate the terms of the series in a different order, as  adding them ''by diagonals''. 
</s>