The elements of the true arithmetic of infinities

&c. and of the fifth series will be 1+31+211+781+2101, &c. the terms in each of these series being formed from the subtraction of the terms from each other in the infinite series of which these latter series are the last terms. The denominators, also, of the fractions, from the expansion the act of developing a fractional expression into a series of which these last terms are produced, are always less than the denominators of the fractions which express the sums of the series by one power.

For 1+1+1+1+1, &c. ad infin. to infinity is evidently equal to the last term of the series 1+2+3+4+5, &c. ad infin. For the sum of 2 of these terms, beginning from the first term, namely 1+1, is equal to the second term of the series 1+2+3+4, &c. The sum of 3 of the terms, beginning from the first term, namely 1+1+1, is equal to 3, or the third term of the series. The sum of 4 of the terms, is equal to 4, or the fourth term of the series, and so on; and therefore the sum of the infinite series 1+1+1+1, &c. will be equal to the last term of the series 1+2+3+4, &c. In like manner 1+3 is equal to 4, the second term of the series 1+4+9+16, &c. 1+3+5 is equal to 9, the third term, and so on. And the same thing will be found to be true in the other series. It is also evident that the terms in the latter are formed from the subtraction of the terms from each other in the former series. Thus, 1+1+1+1, &c. arises from the subtraction of 1 from 2, of 2 from 3, of 3 from 4, &c.; and 1+3+5+7, &c. arises from the subtraction of 1 from 4, of 4 from 9, of 9 from 16, and so on; the first term 1 being excepted. And the like may be shown in the other series. The fractions, likewise, from the expansion of which these last terms are produced, are as follows: The fraction equivalent to the first is $\frac{1}{1-1}$, to the second is $\frac{1+1}{1-2+1}$, to the third is $\frac{1+4+1}{1-3+3-1}$, to the fourth is $\frac{1+11+11+1}{1-4+6-4+1}$, and to the fifth is $\frac{1+26+66+26+1}{1-5+10-10+5-1}$. But the denominator of the fraction $\frac{1}{1-2+1}$, is the second power of the denominator of the fraction $\frac{1}{1-1}$; the denominator of the fraction $\frac{1+1}{1-3+3-1}$, is the third power of 1—1, and the denominator of the fraction $\frac{1+4+1}{1-2+1}$ is only the second power of 1—1. Thus, also, in the third series, the denominator 1—4+6—4—1, is the fourth power of 1—1, but the denominator 1—3+3—1, is only the third power of 1—1; and so of the rest.

COROL. Corollary If these series be supposed to begin from 0, then it is evident that the expressions, equivalent to their last terms, will be $\frac{0+1}{1-1}$, $\frac{0+1+1}{1-2+1}$, $\frac{0+1+4+1}{1-3+3-1}$,