ch1.2.13-negative-powers-infinite-series

Ch 1.2.13 — Of the Resolution of Negative Powers

Summary: Applies the general binomial series to negative integer exponents, converting expressions like $1/(a+b{)}^m$ into infinite power series; verifies each result by multiplication.

Sources: chapter-1.2.13

Last updated: 2026-04-29

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Fractions as Negative Powers (article 370)

Since $1/a=a^{-1}$, the fraction $1/(a+b)$ equals $(a+b{)}^{-1}$, a power of $a+b$ with exponent $-1$. The general binomial series from ch1.2.12-irrational-powers-infinite-series therefore applies. (source: chapter-1.2.13)

$n=-1$: Series for $1/(a+b)$ (article 371)

Substituting $n=-1$ into the general formula, all coefficient fractions equal $-1$:

$\frac{n}{1}=-1,\frac{n-1}{2}=-1,\frac{n-2}{3}=-1,\dots$

The powers of $a$ become $a^{-1},a^{-2},a^{-3},\dots$, giving:

$$\frac{1}{a+b}=\frac{1}{a}-\frac{b}{a^2}+\frac{b^2}{a^3}-\frac{b^3}{a^4}+\frac{b^4}{a^5}-\cdots$$

This is the same geometric series obtained by long division in ch1.2.5-infinite-series, confirming the formula. (source: chapter-1.2.13)

$n=-2$: Series for $1/(a+b{)}^2$ (article 372)

With $n=-2$, the coefficients simplify: $\frac{2\cdot 3\cdots (k+1)}{k!}=k+1$, giving natural-number coefficients:

$$\frac{1}{(a+b{)}^2}=\frac{1}{a^2}-\frac{2b}{a^3}+\frac{3b^2}{a^4}-\frac{4b^3}{a^5}+\frac{5b^4}{a^6}-\cdots$$

(source: chapter-1.2.13)

$n=-3$: Series for $1/(a+b{)}^3$ (article 373)

$$\frac{1}{(a+b{)}^3}=\frac{1}{a^3}-\frac{3b}{a^4}+\frac{6b^2}{a^5}-\frac{10b^3}{a^6}+\frac{15b^4}{a^7}-\frac{21b^5}{a^8}+\frac{28b^6}{a^9}-\cdots$$

The coefficients $1,3,6,10,15,21,28,\dots$ are the triangular numbers. (source: chapter-1.2.13)

$n=-4$: Series for $1/(a+b{)}^4$ (article 373)

$$\frac{1}{(a+b{)}^4}=\frac{1}{a^4}-\frac{4b}{a^5}+\frac{10b^2}{a^6}-\frac{20b^3}{a^7}+\frac{35b^4}{a^8}-\frac{56b^5}{a^9}+\cdots$$

(source: chapter-1.2.13)

General Formula (article 374)

For any positive integer $m$:

$$\frac{1}{(a+b{)}^m}=\frac{1}{a^m}-\frac{mb}{1\cdot a^{m+1}}+\frac{m(m-1)b^2}{1\cdot 2\cdot a^{m+2}}-\frac{m(m-1)(m-2)b^3}{1\cdot 2\cdot 3\cdot a^{m+3}}+\cdots$$

Fractional values of $m$ can also be substituted to express irrational quantities. See binomial-theorem. (source: chapter-1.2.13)

Verification by Multiplication (articles 375–377)

Euler checks each result by multiplying the series back by the original denominator:

• $(a+b{)}^{-1}\cdot (a+b)$: all intermediate terms cancel, leaving 1. ✓
• $(a+b{)}^{-2}\cdot (a+b{)}^2$: three rows of products cancel to leave 1. ✓
• $(a+b{)}^{-2}\cdot (a+b)$: gives exactly the $n=-1$ series. ✓

These verifications are a precursor to termwise manipulation of power series identities. (source: chapter-1.2.13)

Pattern of Coefficients

$m$	Series coefficients
$-1$	$1,-1,1,-1,\dots$ (all ones, alternating)
$-2$	$1,-2,3,-4,5,\dots$ (natural numbers)
$-3$	$1,-3,6,-10,15,\dots$ (triangular numbers)
$-4$	$1,-4,10,-20,35,\dots$ (tetrahedral numbers)

Each row of positive coefficients is the row one step below in pascal-triangle, shifted by the denominator’s power. (source: chapter-1.2.13)

Related pages

• binomial-theorem
• ch1.2.12-irrational-powers-infinite-series
• ch1.2.5-infinite-series
• infinite-series
• pascal-triangle
• powers-and-exponents