The Elements of the True Arithmetic of Infinites

The series also of the aggregates will be 2 / (1-2+1), which, compared to the series of the equals, or 1 / (1-2+1), will be double, and not one and a half times as large.

Again, "If a series of equals be reduced by a series of second-order terms, third-order, fourth-order, etc., the remainders will be two-thirds of the total, three-quarters, four-fifths, etc. But if it is increased in this way, the aggregates will be one and a third, one and a quarter, one and a fifth, etc." original: "Si series æqualium mulctetur serie secundanorum, tertianorum, quartanorum, &c. residua erunt totius duo trientes, tres quadrantes, quatuor quintantes, &c. Sin ita augeatur, erunt aggregata ejusdem sesquitertium, sesquiquartum, sesquiquintum, &c."

For in the series 0+1+4+9+16, etc. = (0+1+1) / (1-3+3-1), the last and greatest term is (0+1+1) / (1-2+1), and 1 / (1-1) is the number of terms; and [1 / (1-1)] × [(0+1+1) / (1-2+1)] = (0+1+1) / (1-3+3-1) = the sum of the series of the greatest term. Then if (0+1+1) / (1-3+3-1) be taken from this sum, the remainder is 0, and not 2/3 of the whole.

Thus, also, in the series 0+1+8+27+64, etc. = (0+1+4+1) / (1-4+6-4+1), the last and greatest term is (0+1+4+1) / (1-3+3-1), which, being multiplied by the number of terms, or 1 / (1-1), is equal to (0+1+4+1) / (1-4+6-4+1). Then (0+1+4+1) / (1-4+6-4+1) - (0+1+4+1) / (1-4+6-4+1) is 0, and not 3/4 of the whole.

PROPOSITION VII.

To demonstrate in various other series of whole numbers that the last term multiplied by the number of terms is equal to the sum of the series.

This is true in the series of terms in arithmetical progression, 1+3+5+7+9, etc. = (1+1) / (1-2+1). For the last term will be equal to (1+1) / (1-1), or 1+2+2+2+2, etc., the addition of the terms of this series being 3, 5, 7, 9, etc. And [1 / (1-1)] × [(1+1) / (1-1)] = (1+1) / (1-2+1), or the sum of the series.