Adding elements of an infinite matrix

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CHANYANG
CHANYANG on 10 Jul 2014
Commented: CHANYANG on 10 Jul 2014
I am studying double series and I want to use matlab to see if a certain double series converges or not.
The double series is \sum \sum a_{nk} , Where [a_{nk}] = [1 -1/2 0 0 0 0 ...; 0 1 -3/4 0 0 ...; 0 0 1 -7/8 0 ...; 0 0 0 1 -15/6 0... ...]
The interesting thing about this double series is that you cannot switch the order of summation. If you add up the rows separately and then add them all together, it's different from adding up the columns and then adding those all up. But I want to add up the elements of square submatrices and increase its size and see if the sum of each submatrix converges as the size grows.
In other words, find a way for matlab to calculate and make a list of desired size such as: 1, (1+-1/2+0+1), (1+-1/2+0+0+1+-3/4+0+0+1), etc, where the first is the sum of the elements of submatrix of dimension 1 and the second number is the sum of the elements of the 2x2 submatrix, etc.

Answers (1)

James Tursa
James Tursa on 10 Jul 2014
You can't use a numeric summation process to see if an infinite sum converges or not. One good example of this is the harmonic series 1 + 1/2 + 1/3 + 1/4 + ... which can easily be proven to diverge to infinity. However, if you code this up in MATLAB (or any numeric language) eventually the partial sum will be too large for the small terms to have any effect on. I.e., the sum will eventually freeze beyond a certain point. So you might think the sum converges because it stops changing, but in fact you are just getting the wrong answer.
If you are trying to just toy around with the various ways of summing to see what you get, then fine. But don't kid yourself that you can prove or even get a reliable hint at convergence this way.
  1 Comment
CHANYANG
CHANYANG on 10 Jul 2014
Thank you for the comment.
Of course, I know that. I was wondering more of how to do something like this in matlab. And I also think I would also know by looking at the partial sums whether the series is going nowhere.

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