continued-fraction-series-equivalence

Continued Fraction ↔ Alternating Series

Summary: Every continued fraction equals an alternating series whose terms are the differences of consecutive convergents, with denominators that are products of consecutive convergent denominators. Conversely, every alternating series can be written as a continued fraction — but only after a free choice of partial denominators, which Euler exploits with several elegant templates. The conversion specialises to Brouncker’s $4/\pi$, the continued fraction for $\log 2$, the continued fractions for $1/(e-1)$ and $\cos 1$, and several parametric families.

Sources: chapter18 (§363–§373).

Last updated: 2026-05-11

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Continued fraction → alternating series (§363–§366)

Let $P,Q,R,S,T,\dots$ denote the successive denominators of the convergents (after the trivial $1/0$ and $a/1$), so $P=b$, $Q=bc+\beta$, $R=bcd+\beta d+\gamma b$, $S=bcde+\beta de+\gamma be+\delta bc+\beta \delta$, etc. By the convergent recurrence these satisfy

$Q=Pc+\beta ,R=Qd+\gamma P,S=Re+\delta Q,T=Sf+ϵR,\dots$

Subtracting consecutive convergents (§363) gives a telescoping identity that Euler writes as

$\boxed{z=a+\frac{\alpha }{P}-\frac{\alpha \beta }{PQ}+\frac{\alpha \beta \gamma }{QR}-\frac{\alpha \beta \gamma \delta }{RS}+\frac{\alpha \beta \gamma \delta ϵ}{ST}-\cdots }$

So the value of the continued fraction is an alternating series whose $k$-th term has numerator $\alpha \beta \gamma \cdots$ ($k$ factors of partial-numerator data) and denominator $P_{k-1}{P}_k$ (the product of two consecutive convergent denominators).

If the partial numerators are all $1$ and the partial denominators $a,b,c,\dots$ are positive integers (the simple form), this series converges very rapidly — its terms decrease at least geometrically.

Alternating series → continued fraction (§365–§368)

Given $x=A-B+C-D+E-\cdots$, Euler matches term by term against the §363 expansion:

$\frac{\alpha }{P}=A,\frac{\alpha \beta }{PQ}=B,\frac{\alpha \beta \gamma }{QR}=C,\frac{\alpha \beta \gamma \delta }{RS}=D,\dots$

Solving sequentially:

$\alpha =AP,\beta =\frac{BQ}{A},\gamma =\frac{CR}{BP},\delta =\frac{DS}{CQ},ϵ=\frac{ET}{DR},\dots$

Since the convergent denominators $P,Q,R,S,\dots$ depend on the partial denominators $b,c,d,e,\dots$ which are free, this gives a one-parameter family per level. Euler exploits the freedom to clear fractions.

Template I — $A,B,C,\dots$ integers (§368)

Set $b=1$, $c=A-B$, $d=B-C$, $e=C-D$, $f=D-E$, $\dots$. Then

$\alpha =A,\beta =B,\gamma =AC,\delta =BD,ϵ=CE,\dots$

and the continued fraction is

$x=\frac{A}{1+\frac{B}{A-B+\frac{AC}{B-C+\frac{BD}{C-D+\frac{CE}{D-E+\cdots }}}}}$

Template II — reciprocal terms (§369)

For $x=1/A-1/B+1/C-1/D+\cdots$, set $b=A$, $c=B-A$, $d=C-B$, $e=D-C$, $\dots$. Then $\alpha =1$, $\beta =A^2$, $\gamma =B^2$, $\delta =C^2$, $\dots$ and

$x=\frac{1}{A+\frac{A^2}{B-A+\frac{B^2}{C-B+\frac{C^2}{D-C+\cdots }}}}.$

The two famous specialisations:

• $A=1,B=2,C=3,\dots$ gives the [[continued-fraction-for-log-2|continued fraction for $\log 2$]] (Example I).
• $A=1,B=3,C=5,\dots$ gives [[brouncker-formula|Brouncker’s continued fraction for $4/\pi$]] (Example II).

A parametric example (III): $A=m,B=m+n,C=m+2n,\dots$ gives the continued fraction for $\sum (-1{)}^k/(m+kn)$. Example IV reuses §178 to convert $\pi \cos (m\pi /n)/(n\sin (m\pi /n))$ into a continued fraction.

Template III — products in denominators (§370)

For $x=1/A-1/(AB)+1/(ABC)-1/(ABCD)+\cdots$, set $b=A$, $c=B-1$, $d=C-1$, $e=D-1$, $\dots$. Then $\alpha =1$, $\beta =A$, $\gamma =B$, $\delta =C$, $\dots$ and

$x=\frac{1}{A+\frac{A}{B-1+\frac{B}{C-1+\frac{C}{D-1+\cdots }}}}.$

The two named specialisations (§370):

• $1/e=1-1+1/2-1/6+1/24-\cdots$ gives $\frac{1}{e-1}=\frac{1}{1+\frac{2}{2+\frac{3}{3+\frac{4}{4+\frac{5}{5+\cdots }}}}}.$
• $\cos 1=1-1/2+1/(2\cdot 12)-1/(2\cdot 12\cdot 30)+\cdots$ gives $\cos 1=\frac{1}{1+\frac{1}{1+\frac{2}{11+\frac{12}{29+\frac{30}{55+\cdots }}}}}.$

Templates IV–V — power-series and z-product variants (§371–§373)

For $x=A-Bz+C{z}^2-D{z}^3+\cdots$, set $b=1$, $c=A-Bz$, $d=B-Cz$, $\dots$, giving partial numerators $\alpha =A$, $\beta =Bz$, $\gamma =ACz$, $\delta =BDz$, $\dots$ The more general template (§372) handles $x=A/L-By/(Mz)+C{y}^2/(N{z}^2)-\cdots$, and (§373) the product-form $x=A/L-ABy/(LMz)+ABC{y}^2/(LMN{z}^2)-\cdots$.

The asymmetry of the conversion (§374–§375)

Series-to-continued-fraction always works; the converse is harder. Euler shows that the simple periodic CF $1/(2+1/(2+\cdots ))$ produces (via the §363 telescoping)

$x=\frac{1}{2}-\frac{1}{2\cdot 5}+\frac{1}{5\cdot 12}-\frac{1}{12\cdot 29}+\cdots ,$

a strongly convergent series whose value is not visible from the series — but is immediately seen from the CF as $\sqrt{2}-1$ by the §376 quadratic-equation trick. So continued fractions sometimes know more than their associated series do.

Related pages

• chapter-18-on-continued-fractions
• continued-fraction
• convergents-of-a-continued-fraction
• brouncker-formula — Template II at $A,B,C,\dots =1,3,5,\dots$
• continued-fraction-for-log-2 — Template II at $A,B,C,\dots =1,2,3,\dots$
• continued-fraction-for-e — Template III at $A,B,C,\dots =1,2,3,\dots$ (after a one-term shift)
• logarithmic-series — source of the $\log 2$ alternating series
• arctangent-series — source of the $\pi /4$ alternating series
• cotangent-partial-fraction — source of the §370 Example IV trig series