How to sum an infinite matrix?

Question

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.

Answer 1

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 $=4$

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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''.

Answer 2

Notice that the second matrix you come up with is not absolutely convergent: you have an infinite number of $2$'s whose sum is $\infty$, and same with an infinite number of $-1$'s. So you can't just rearrange terms like that, and if you do, you get a wrong answer. What you need to do is not write $1$ as $2-1$, but just as $1$. And then $\frac{1}{2}$ is not $2-1-\frac{1}{2}$, but just $\frac{1}{2}$. If you do this, the sequence should be familiar and you should be able to easily find the sum.

Answer 3

If you compute the sum of each row, you'll notice that it's 2 * (1/2)^n (n going from 0 to infinity). Which is iteself is an infinite series that trivially sums to 4.

Where I think you are going wrong: When summing a matrix, you have to add to it in a square fashion. So, a row either exists or does not. So while you can consider the first row of having a value of (2 - 1 - 1/2 -1/4) once the 4th row is considered, you have to add the value of the 2nd row (2- 1 - 1/2), 3rd row ( 2 -1 ) and 4th row ( 2). When you want to shrink those values, you do that by adding the fifth row, but that adds its own 2 into the mix...

Answer 4

There is a theorem that says that a double series is summable if

$$\sup_{n\in\Bbb N}\sum_{j,k=0}^n|a_{j,k}|<\infty$$

Your method was wrong because the triangular matrix that you derived is not summable as a double series, that is

$$\sup_{n\in\Bbb N}\sum_{j,k=0}^n|a_{j,k}|\ge\sup_{n\in\Bbb N}\sum_{j=0}^n|a_{j,0}|=\sup_{n\in\Bbb N}\sum_{j=0}^n 2=\infty$$

That is: you can see that if you change the order in the sum of the triangular matrix the sum will be different, hence this matrix is not summable.

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Observe that the sum of the original matrix is equivalent to the sum of this double series

$$\sum_{k=0}^\infty\sum_{j=0}^\infty\frac1{2^{j+k}}$$

what is summable because

$$\begin{align}\sum_{j,k=0}^\infty\left(\frac12\right)^{j+k}&=\lim_{n\to\infty}\sum_{j,k=0}^n\left(\frac12\right)^{j+k}\\&=\lim_{n\to\infty}\sum_{m=0}^n(m+1)\left(\frac12\right)^m,\qquad\text{renaming }m=j+k\\&=\sum_{m=0}^\infty(m+1)\left(\frac12\right)^m\\&=\left[\sum_{m=0}^\infty(m+1)x^m\right]_{x=1/2}\\&=\left[\sum_{m=0}^\infty \partial_x x^{m+1}\right]_{x=1/2}\\&=\left[\partial_x\sum_{m=0}^\infty x^{m+1}\right]_{x=1/2},\qquad\text{because the geometric series is analytic in }|x|<1\\&=\left[\partial_x\frac{x}{1-x}\right]_{x=1/2},\qquad \text{if }|x|<1\\&=\left[\frac{1}{(1-x)^2}\right]_{x=1/2}=4<\infty\end{align}$$

because there are $m+1$ ways to sum up to $m$ using two non-negative integers, that is

$$m+1=\binom{m+2-1}{2-1}$$

Check it here or here as weak compositions of $m$ with two non-negative integers.