Page:Philosophical Transactions

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And that therefore in the first series half the first term is greater than the sum of the two next, and half this sum of the second and third greater than the sum of the four next, and half the sum of those four greater than the sum of the next eight, &c. in infinitum. For 1/2dD = br + bn, but bn > fG, therefore 1/2dD > br + fG, &c. And in the second series half the first term is less then the sum of the two next, and half this sum less then the sum of the four next, &c. in infinitum.

That the frist series are the even terms, viz. the 2^d, 4^th, 6^th, 8^th, 10^th, &c. and the second, the odd, viz. the 1^st, 3^rd, 5^th, 7^th, 9^th, &c. of the following series, viz. ${\tfrac {1}{1\times 2}}+{\tfrac {1}{2\times 3}}+{\tfrac {1}{3\times 4}}+{\tfrac {1}{4\times 5}}+{\tfrac {1}{5\times 6}}+{\tfrac {1}{6\times 7}}+$ &c. in inifinitum = 1. Whereof a being put for the number of terms taken at pleasure, ${\tfrac {1}{a+a}}$ is the last, ${\tfrac {a}{a+1}}$ is the sum of all those terms from the beginning, and ${\tfrac {1}{a+1}}$ the sum of the rest to the end.

That 1/4 of the first terme in the third series is less than the sum of the two next, and a quarter of this sum, less than the sum of the four next, and one fourth of this last sum less than the next eight, I thus demonstrate.

Let a = the 3^d or last number of any term of the first Column, viz. of Divisors,

${\frac {1}{a\times {\underline {a-1}}\times {\underline {a-2}}}}={\frac {1}{a^{3}-3a^{2}+2^{a}}}={\frac {16a^{3}-48a^{2}+56a-24}{16a^{6}-96a^{5}+232a^{4}-288a^{3}+184a^{2}-48a}}=A$

${\frac {1}{{\underline {2a}}\times {\underline {2a-1}}\times {\underline {2a-2}}}}={\frac {1}{8a^{3}-12a^{2}+4^{a}}}$	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \ \end{matrix}}\right\}\,}$ $={\frac {16a^{3}-48a^{2}+56a-24}{16a^{6}-384a^{5}+880a^{4}-960a^{3}+498a-96}}=B$
${\frac {1}{{\underline {2a-2}}\times {\underline {2a-3}}\times {\underline {2a-4}}}}={\frac {1}{8a^{3}-36a^{2}+52a-24}}$

${\frac {64a^{6}-384a^{5}+928a^{4}-1152a^{3}+736a^{2}-102a}{64a^{6}-384a^{5}+880a^{4}-960a^{2}-96a}}\times {\tfrac {1}{4}}A<B$ .

And $48a^{4}-192a^{3}+240a^{2}-96a=$ Excess of the Numerator above Denomin.

But —— The affirm.	$>$	the Negat.	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \ \end{matrix}}\right\}\,}$ if a > 2.
That is, $48a^{3}+240a^{2}$	$>$	$192a^{3}+96a$
Because $a^{3}+5a^{2}$	$>$	$4a^{3}+2a$
$a^{2}+5a$	$>$	$4a^{2}+2$

Therefore $B>{\tfrac {1}{4}}A$ .

Therefore 14 of any number of A; or Terms, is less than their so many respective B. that is, than twice so many of the next Terms., Quod, &c.

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