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.