What about magic squares that use 1/N for all N and use no other numbers?

They exist. Suppose we want a sum of K > 1. We use a generalisation of the `back-and-forth' argument to construct such a bijection between {1, 1/2, 1/3, ...} and the grid N x N. Specifically, create an enumeration of the following [countable set of] assertions: 1. "The number 1/N has been placed." 2. "The square (x,y) has been occupied." 3. "The nth line has a sum greater than K - p/q." Here `line' means `row, column or main diagonal'. Begin with an initially empty grid. Now, for each assertion, place unused numbers in unoccupied squares until the assertion is true, ensuring that at no point do we let any line have a sum >= K. We can always do this, since at any point only finitely many numbers have been placed, and only finitely many squares have been occupied. In particular, for each form of assertion, do the following: 1. Place 1/N on a square such that all lines incident with the square were previously unoccupied. 2. Choose N sufficiently large that 1/N is strictly less than the `deficiencies' of the lines incident with (x,y) (where `deficiency' is defined to be the difference between the current sum and the target K). 3. Repeatedly place the largest unused number less than the deficiency on an unoccupied square on the nth line (such that all other lines incident with this square were hitherto unoccupied). This process will terminate since 1/1 + 1/2 + 1/3 + 1/4 + ... diverges. Now consider the limit of this process (which is well-defined, since it's an increasing union of injective partial functions between {1, 1/2, 1/3, ...} and N x N). This is clearly a bijection, so every square is occupied and every number is used. Moreover, no sum exceeds K (since otherwise we can find a finite subset of the line which exceeds K, but we avoided doing this). Also, no sum is less than K (since otherwise we could find p/q such that the sum is less than K - p/q, but we asserted that each line will have sum exceeding K - p/q). Consequently, every sum is equal to K. The result follows. QED. Sincerely, Adam P. Goucher

Consider:

1 1/2 1/6 1/7 1/15 1/16 ... 1/3 1/5 1/8 1/14 1/17 ... 1/4 1/9 1/13 1/18 ... 1/10 1/12 1/19 ... 1/11 1/20 ... 1/21 ... ...

in which I just wrote each successive reciprocal on successive cells on successive diagonals, such that you could draw a curve through them in succession without lifting the pen. At least every row and every column converges, as does the main diagonal. Of course they converge to different values. (At least I assume no two sums are the same. Can anyone prove or disprove it?)

Is it possible to make an arrangement in which every row and every column converge if you use *all* rationals between 0 and 1, rather than just the ones with a numerator of 1? If not, what about in higher dimensions?

I expect it would be very easy to find an infinite magic square if negative values are allowed. So lets keep it positive.

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