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.


The ''triangular array'' is not really an array but a column of values:

$ 2 $






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''.

  • $\begingroup$ But there is a diagonal... why can't I use it? $\endgroup$ – GambitSquared Jun 27 '17 at 20:01
  • $\begingroup$ I added something to my answer. I hope it's useful :) $\endgroup$ – Emilio Novati Jun 27 '17 at 20:14
  • $\begingroup$ Why can we not rearrange or associate the terms? $\endgroup$ – GambitSquared Jun 27 '17 at 21:01
  • $\begingroup$ We can rearrange the terms only if the series is absolutely convergent,: en.wikipedia.org/wiki/Absolute_convergence $\endgroup$ – Emilio Novati Jun 27 '17 at 21:05

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.

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


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


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

  • 2
    $\begingroup$ This is a good answer. A very similar argument is that the infinite sum that is interpreted as a triangular matrix is not absolutely convergent, and as such it is not infinitely commutative. $\endgroup$ – Eddy Jun 27 '17 at 20:49

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.


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...

  • $\begingroup$ I see that... but my question is really: why doesn't my method yield the right answer? What is fundamentally wrong with the method? $\endgroup$ – GambitSquared Jun 27 '17 at 20:02

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