Your nineteenth proposition is this other lemma: “_In a series, or a row, of quantities, beginning from a point, or cypher, and proceeding according to the order of the square numbers, as_ 0, 1, 4, 9, 16, _&c. to find what proportion the whole series hath to so many times the greatest_.” And you conclude “_the proportions to be that of 1 to 3_.” Which is false, as you shall presently see. First, let the series of squares with the prefixed cypher, and under every one of them the greatest 4 be (0 . 1 . 4)/(4 . 4 . 4). And you have for the sum of the squares 5, and for thrice the greatest 12, the third part whereof is 4. But 5 is greater than 4, by 1, that is, by one twelfth of 12; which quantity is somewhat, let it be called A. Again, let the row of squares be lengthened one term further, and the greatepm divst set under every one of them as (0 . 1 . 4 . 9)/(9 . 9 . 9 . 9). The sum of the squares is 14, and the sum of four times the greatest is 36, whereof the third part is 12. But 14 is greater than 12 by two unities, that is, by two twelfths of 12, that is, by 2 A. The difference therefore between the sum of the squares, and the sum of so many times the greatest square, is greater, when the cypher is followed by three squares, than when by but two. Again, let the row have five terms, as in these numbers (0 . 1 . 4 . 9 . 16)/(16 . 16 . 16 . 16 . 16) with the greatest five times described, and the sum of the squares will be 30, the sum of all the greatest will be 80. The third part whereof is 26(2)/(3). But 30 is greater than 26(2)/(3) by 3(1)/(3), that is, by three twelfths of twelve, and (1)/(3) of a twelfth, that is, by 3(1)/(3) A. Likewise in the series continued to six places with the greatest six times subscribed, as ( 0 . 1 . 4 . 9 . 16 . 25)/(25 . 25 . 25 . 25 . 25 . 25) the sum of the squares is 55, and the sum of the greatest six times taken is 150, the third part whereof is 50. But 55 is greater than 50 by 5, that is, by five-twelfths of 12, that is by 5 A. And so continually as the row groweth longer, the excess also of the aggregate of the squares above the third part of the aggregate of so many times the greatest square, growing greater. And consequently if the number of the squares were infinite, their sum would be so far from being equal to the third part of the aggregate of the greatest as often taken, as that it would be greater than it by a quantity greater than any that can be given or named.