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

"If a series of squares, whose sides or roots are in arithmetical progression, beginning with a cypher, be infinitely continued, the last term being multiplied into the number of terms will be triple the sum of all the series.

"If a series of cubes, whose roots are in arithmetical progression, beginning with a cypher, be infinitely continued, the last term being multiplied into the number of terms will be quadruple the sum of all the series.

"If a series of biquadrats fourth powers of a number, whose roots are in arithmetical progression, beginning with a cypher, be infinitely continued, the last term multiplied into the number of terms will be quintuple the sum of all the series."

For by the preceding proposition, the last term in each of these series multiplied by the number of terms is exactly equal to the sum of the series.

COROL. Corollary Hence, as the whole of the Arithmetic of Infinites of Dr. Wallis is founded on the above false propositions, no part of that arithmetic is to be considered as demonstrative proven by logical necessity; and such conclusions in it, as may happen to be true, are not legitimately deduced.

---

PROP. VI.

The following propositions, also, in the Arithmetic of Infinites, are false. See the 84th and 85th pages of Dr. Wallis's Treatise.

"If a series of equals be respectively subtracted from a series of first powers original Latin: "Si series æqualium serie primanorum respective mulctetur" (for instance, if the first term of the latter is taken from the first of the former, the second from the second, etc.) the remainders will be half of the whole; but if it is so increased, the series of sums will be one and a half times original Latin: "sesquialtera" the series of equals shown."

For the greatest and last term of the series $0+1+2+3+4$, &c. $= \frac{0+1}{1-2+1}$, is equal to $\frac{0+1}{1-1}$, which being multiplied by the number of terms, gives $\frac{0+1}{1-2+1}$. And if $\frac{1}{1-2+1}$ be taken from this, the remainder will be 0, and not $\frac{1}{2}$ of the whole.