THE ELEMENTS OF THE TRUE ARITHMETIC OF INFINITES, &c.

The series also of the aggregates the total sums formed by addition will be $\frac{2}{1-2+1}$, which to the series of the equals, or $\frac{1}{1-2+1}$, will be double, and not sesquialter one and a half.

Again, "If a series of equals is diminished by a series of second powers, third powers, fourth powers, etc. original Latin: "Si series æqualium mulctetur serie secundanorum, tertianorum, quartanorum, &c.", the remainders will be two-thirds, three-fourths, four-fifths, etc. original Latin: "duo trientes, tres quadrantes, quatuor quintantes, &c." of the whole. But if it is so increased, the sums will be one and a third, one and a fourth, one and a fifth, etc. original Latin: "sesquitertium, sesquiquartum, sesquiquintum, &c." of the same."

For in the series $0+1+4+9+16$, &c. $=\frac{0+1+1}{1-3+3-1}$, the last and greatest term is $\frac{0+1+1}{1-2+1}$, and $\frac{1}{1-1}$ is the number of terms; and $\frac{1}{1-1} \times \frac{0+1+1}{1-2+1} = \frac{0+1+1}{1-3+3-1} =$ the sum of the series of the greatest term. Then if $\frac{0+1+1}{1-3+3-1}$ be taken from this sum, the remainder is 0, and not $\frac{2}{3}$ of the whole.

Thus, also, in the series $0+1+8+27+64$, &c. $=\frac{0+1+4+1}{1-4+6-4+1}$, the last and greatest term is $\frac{0+1+4+1}{1-3+3-1}$, which being multiplied by the number of terms, or $\frac{1}{1-1}$, is equal to $\frac{0+1+4+1}{1-4+6-4+1}$. Then $\frac{0+1+4+1}{1-4+6-4+1} - \frac{0+1+4+1}{1-4+6-4+1}$ is 0, and not $\frac{3}{4}$ of the whole.

PROP. 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, $1+3+5+7+9$, &c. $=\frac{1+1}{1-2+1}$. For the last term will be equal to $\frac{1+1}{1-1}$, or $1+2+2+2+2$, &c. the addition of the terms of this series being $3, 5, 7, 9$, &c. And $\frac{1}{1-1} \times \frac{1+1}{1-1} = \frac{1+1}{1-2+1}$, or the sum of the series.