An infinite magic square would look something like this: 1/2 1/4 1/8 1/16 1/32 1/64 ... 2/3 2/9 2/27 2/81 2/243 2/729 ... 3/4 3/16 3/64 3/256 3/1024 3/4096 ... 4/5 4/25 4/125 4/625 4/3125 4/15625 ... 5/6 5/36 5/216 5/1296 5/7776 5/46656 ... ... ... ... ... ... ... R/(R+1)^C No number would appear more than once, and every row, every column, and every main diagonal would sum to (i.e. converge to) the same sum. (I know what you're thinking: There can't be an infinite magic square since there's only one main diagonal. But I assert that it's valid if *every* main diagonal works, even if there's only one main diagonal.) In my arrangement above, no number appears more than once, and every row has the same sum. Unfortunately the columns and the main diagonal don't work. Indeed, the first two columns don't even converge. If such a magic square can't be found, or can be proven not to exist, perhaps a semi-magic square can be found. Maybe even by permuting each row of my above non-solution. For instance what about arrangements that always move the largest, left-most, term to the diagonal? What about magic squares that use 1/N for all N and use no other numbers? 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.