brouncker-formula

Brouncker’s Continued Fraction for $4/\pi$

Summary: Applying the §369 reciprocal-series template to the Leibniz series $\pi /4=1-1/3+1/5-1/7+\cdots$ produces William Brouncker’s 1655 continued fraction $4/\pi =1+1^2/(2+3^2/(2+5^2/(2+7^2/(2+\cdots ))))$ — the first continued fraction for $\pi$ in the history of mathematics, recovered here by Euler as a particular case of his general series-to-CF dictionary.

Sources: chapter18 (§369 Example II).

Last updated: 2026-05-11

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Derivation

The Leibniz series for $\pi /4$ has the form $1/A-1/B+1/C-1/D+\cdots$ with $A=1,B=3,C=5,D=7,\dots$. Applying Template II of §369 — set $b=A=1$, $c=B-A=2$, $d=C-B=2$, $e=D-C=2$, $\dots$ — gives partial numerators $\alpha =1,\beta =A^2=1,\gamma =B^2=9,\delta =C^2=25,ϵ=D^2=49,\dots$ and partial denominators $b=1,c=d=e=\cdots =2$. Therefore

$\frac{\pi }{4}=\frac{1}{1+\frac{1}{2+\frac{9}{2+\frac{25}{2+\frac{49}{2+\cdots }}}}}.$

Inverting (Euler’s preferred form, §369 Example II) eliminates the leading $1/(1+\cdots )$:

$\boxed{\frac{4}{\pi }=1+\frac{1^2}{2+\frac{3^2}{2+\frac{5^2}{2+\frac{7^2}{2+\frac{9^2}{2+\cdots }}}}}}$

Numerators are the squares of the odd integers; all partial denominators equal $2$ from the second level onward.

Historical note

Euler attributes the identity directly: “this is the expression first found by BROUNCKER as a quadrature of the circle.” William Brouncker (1620–1684), first president of the Royal Society, communicated it to John Wallis around 1655 in connection with Wallis’s Arithmetica infinitorum; it was Wallis who in turn proved it equivalent to his own infinite product $\pi /2=(2\cdot 2\cdot 4\cdot 4\cdots )/(1\cdot 3\cdot 3\cdot 5\cdots )$. The Brouncker continued fraction was the first ever written down for $\pi$, predating Euler’s series-CF dictionary by nearly a century.

Convergence

Although the form is elegant, convergence is slow — comparable to the Leibniz series itself. The $n$-th convergent has error of order $1/n$, not $1/n^2$ or geometric, because the partial numerators $(2k+1{)}^2$ grow at exactly the rate that cancels the geometric decay one would expect from a simple CF with bounded partial numerators. So Brouncker’s formula is more aesthetic than computational; for fast computation of $\pi$ Euler had already given Machin’s formula $\pi /4=\arctan (1/2)+\arctan (1/3)$ in chapter 8, and §382 will give the convergents of $\pi$ directly via the Euclidean-algorithm method.

Related pages

• continued-fraction-series-equivalence — the §369 reciprocal-series template Euler applies here
• arctangent-series — source of the Leibniz $\pi /4$ series
• wallis-product — Wallis’s product, which Brouncker’s CF was derived in connection with
• machin-like-formula — much faster way of computing $\pi$
• best-rational-approximations — §382 derives a different (faster) CF for $\pi$ by the Euclidean method
• chapter-18-on-continued-fractions