ch1.2.5-infinite-series

Chapter V – Of the Resolution of Fractions into Infinite Series

Summary: Shows that any fraction whose divisor does not divide the dividend can be expanded, via repeated long division, into a power series with infinitely many terms; illustrates this for geometric and alternating series and confirms that such infinite series can have determinate finite values.

Sources: chapter-1.2.5

Last updated: 2026-04-26

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§289–291: Repeated Division of $1/(1-a)$

Euler observes that when a division is inexact the result is a fraction, but nothing prevents continuing the division indefinitely. Dividing $1$ by $1-a$ step by step produces a growing quotient with a shrinking remainder: (source: chapter-1.2.5)

Step	Quotient	Remainder
1	$1+\frac{a}{1-a}$	$a$
2	$1+a+\frac{a^2}{1-a}$	$a^2$
3	$1+a+a^2+\frac{a^3}{1-a}$	$a^3$

In each case the quotient plus the fractional remainder equals the original fraction:

$1+a+a^2+\frac{a^3}{1-a}=\frac{1-a^3}{1-a}+\frac{a^3}{1-a}=\frac{1}{1-a}.$

(source: chapter-1.2.5)

§292: The Infinite Geometric Series

Continuing without limit gives the geometric series:

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

Euler states that there are “sufficient grounds” to maintain that the value of this infinite series equals $\frac{1}{1-a}$. (source: chapter-1.2.5)

§293–297: Numerical Examples

$a=1$: The series $1+1+1+1+\cdots$ must equal $\frac{1}{1-1}=\frac{1}{0}$, confirming Euler’s earlier result that $\frac{1}{0}=\infty$. See infinity. (source: chapter-1.2.5)

$a=2$: The series $1+2+4+8+\cdots$ must equal $\frac{1}{1-2}=-1$, which seems absurd. Euler explains the paradox: if one stops at any term one must adjoin the remaining fractional tail; for example, stopping at 64 adds $\frac{128}{1-2}=-128$, giving $127-128=-1$ as required. (source: chapter-1.2.5)

$a=\frac{1}{2}$: $\frac{1}{1-\frac{1}{2}}=2$, so $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots =2$. Taking more terms the partial sums approach 2 with an error halving each time. (source: chapter-1.2.5)

$a=\frac{1}{3}$: $1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots =\frac{3}{2}$; error shrinks threefold with each term. (source: chapter-1.2.5)

$a=\frac{2}{3}$: $1+\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\cdots =3$. (source: chapter-1.2.5)

$a=\frac{1}{4}$: $1+\frac{1}{4}+\frac{1}{16}+\cdots =\frac{4}{3}$. (source: chapter-1.2.5)

When $|a|<1$ partial sums converge steadily to $\frac{1}{1-a}$; when $|a|\ge 1$ the remainder term does not vanish and the series in isolation does not converge in the modern sense.

§298–301: The Alternating Series $1/(1+a)$

Dividing 1 by $1+a$ in the same way yields the alternating geometric series: (source: chapter-1.2.5)

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

$a=1$ (Grandi’s series): $\frac{1}{2}=1-1+1-1+\cdots$. Euler notes the apparent contradiction: stopping at $-1$ gives 0, stopping at $+1$ gives 1. His resolution is that since the series never terminates, the sum lies between 0 and 1, and therefore equals $\frac{1}{2}$. (source: chapter-1.2.5)

$a=\frac{1}{2}$: $1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots =\frac{2}{3}$; partial sums alternate above and below with halving error. (source: chapter-1.2.5)

$a=\frac{1}{3}$: $1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\cdots =\frac{3}{4}$. (source: chapter-1.2.5)

§302: Alternative Expansion in Negative Powers

By treating the divisor as $a+1$ and dividing 1 by $a$ first, a second series emerges: (source: chapter-1.2.5)

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

Setting $a=2$ gives $\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\cdots =\frac{1}{3}$. (source: chapter-1.2.5)

§303: General Fraction $c/(a+b)$

For the general fraction with two-term denominator: (source: chapter-1.2.5)

$\frac{c}{a+b}=\frac{c}{a}-\frac{bc}{a^2}+\frac{b^{2}c}{a^3}-\frac{b^{3}c}{a^4}+\cdots$

Example with $a=10,b=1,c=11$: $1=\frac{11}{10}-\frac{11}{100}+\frac{11}{1000}-\frac{1}{10000}+\cdots$ Partial sums alternate between slightly more and slightly less than 1, with errors $\frac{1}{10},\frac{1}{100},\frac{1}{1000},\dots$ (source: chapter-1.2.5)

§304: Trinomial Divisor

Dividing 1 by the three-term expression $1-a+a^2$ gives a series with a period-4 sign pattern: (source: chapter-1.2.5)

$\frac{1}{1-a+a^2}=1+a-a^3-a^4+a^6+a^7-\cdots$

Setting $a=1$ recovers $1=1+1-1-1+1+1-\cdots$, which contains two copies of Grandi’s series $\frac{1}{2}+\frac{1}{2}=1$. (source: chapter-1.2.5)

For $a=\frac{1}{2}$: $\frac{4}{3}=1+\frac{1}{2}-\frac{1}{8}-\frac{1}{16}+\frac{1}{64}+\cdots$ (source: chapter-1.2.5)

§305: Conclusion

Euler remarks that the method of resolving fractions into infinite series is of the “greatest utility.” He emphasises two points: (source: chapter-1.2.5)

1. An infinite series, though it never terminates, may have a determinate value.
2. This branch of mathematics has led to “inventions of the utmost importance.”

Practice Questions

Five exercises in expansion (source: chapter-1.2.5):

Fraction	Infinite series
$\frac{ax}{a-x}$	$x+\frac{x^2}{a}+\frac{x^3}{a^2}+\frac{x^4}{a^3}+\cdots$
$\frac{b}{a+x}$	$\frac{b}{a}\left(1-\frac{x}{a}+\frac{x^2}{a^2}-\frac{x^3}{a^3}+\cdots \right)$
$\frac{a^2}{x+b}$	$\frac{a^2}{x}\left(1-\frac{b}{x}+\frac{b^2}{x^2}-\frac{b^3}{x^3}+\cdots \right)$
$\frac{1+x}{1-x}$	$1+2x+2x^2+2x^3+2x^4+\cdots$
$\frac{a^2}{(a+x{)}^2}$	$1-\frac{2x}{a}+\frac{3x^2}{a^2}-\frac{4x^3}{a^3}+\cdots$

Related pages

• infinite-series
• fractions
• infinity
• ch1.2.4-division-compound-quantities
• division-of-compound-quantities
• ch1.1.7-fractions-in-general