Continued fraction

Consider, for example, the rational number 415/93, which is around 4.4624. As a first approximation, start with 4, which is the integer part; 415/93 = 4 + 43/93. The fractional part is the reciprocal of 93/43 which is about 2.1628. Use the integer part, 2, as an approximation for the reciprocal to obtain a second approximation of 4 + 1/2 = 4.5. Now, 93/43 = 2 + 7/43; the remaining fractional part, 7/43, is the reciprocal of 43/7, and 43/7 is around 6.1429. Use 6 as an approximation for this to obtain 2 + 1/6 as an approximation for 93/43 and 4 + 1/2 + 1/6, about 4.4615, as the third approximation. Further, 43/7 = 6 + 1/7. Finally, the fractional part, 1/7, is the reciprocal of 7, so its approximation in this scheme, 7, is exact (7/1 = 7 + 0/1) and produces the exact expression 4 + 1/2 + 1/6 + 1/7 for 415/93.

The expression 4 + 1/2 + 1/6 + 1/7 is called the continued fraction representation of 415/93. This can be represented by the abbreviated notation 415/93 = [4; 2, 6, 7]. (It is customary to replace only the first comma by a semicolon.) Some older textbooks use all commas in the (n + 1)-tuple, for example, [4, 2, 6, 7].[3][4]

If the starting number is rational, then this process exactly parallels the Euclidean algorithm applied to the numerator and denominator of the number. In particular, it must terminate and produce a finite continued fraction representation of the number. The sequence of integers that occur in this representation is the sequence of successive quotients computed by the Euclidean algorithm. If the starting number is irrational, then the process continues indefinitely. This produces a sequence of approximations, all of which are rational numbers, and these converge to the starting number as a limit. This is the (infinite) continued fraction representation of the number. Examples of continued fraction representations of irrational numbers are:

• √19 = [4;2,1,3,1,2,8,2,1,3,1,2,8,...] (sequence A010124 in the OEIS). The pattern repeats indefinitely with a period of 6.
• e = [2;1,2,1,1,4,1,1,6,1,1,8,...] (sequence A003417 in the OEIS). The pattern repeats indefinitely with a period of 3 except that 2 is added to one of the terms in each cycle.
• π = [3;7,15,1,292,1,1,1,2,1,3,1,...] (sequence A001203 in the OEIS). No pattern has ever been found in this representation.
• Φ = [1;1,1,1,1,1,1,1,1,1,1,1,...] (sequence A000012 in the OEIS). The golden ratio, the irrational number that is the "most difficult" to approximate rationally (see § A property of the golden ratio φ below).
• γ = [0;1,1,2,1,2,1,4,3,13,5,1,...] (sequence A002852 in the OEIS). The Euler–Mascheroni constant, which is expected but not known to be irrational, and whose continued fraction has no apparent pattern.

Continued fractions are, in some ways, more "mathematically natural" representations of a real number than other representations such as decimal representations, and they have several desirable properties:

• The continued fraction representation for a real number is finite if and only if it is a rational number. In contrast, the decimal representation of a rational number may be finite, for example 137/1600 = 0.085625, or infinite with a repeating cycle, for example 4/27 = 0.148148148148...
• Every rational number has an essentially unique simple continued fraction representation. Each rational can be represented in exactly two ways, since [a₀;a₁,... a_n−1,a_n] = [a₀;a₁,... a_n−1,(a_n−1),1]. Usually the first, shorter one is chosen as the canonical representation.
• The simple continued fraction representation of an irrational number is unique. (However, additional representations are possible when using generalized continued fractions; see below.)
• The real numbers whose continued fraction eventually repeats are precisely the quadratic irrationals.[5] For example, the repeating continued fraction [1;1,1,1,...] is the golden ratio, and the repeating continued fraction [1;2,2,2,...] is the square root of 2. In contrast, the decimal representations of quadratic irrationals are apparently random. The square roots of all (positive) integers that are not perfect squares are quadratic irrationals, and hence are unique periodic continued fractions.
• The successive approximations generated in finding the continued fraction representation of a number, that is, by truncating the continued fraction representation, are in a certain sense (described below) the "best possible".

A (generalized) continued fraction is an expression of the form

${\displaystyle a_{0}+{\cfrac {b_{1}}{a_{1}+{\cfrac {b_{2}}{a_{2}+{\cfrac {b_{3}}{a_{3}+{_{\ddots }}}}}}}}}$

where a_i and b_i can be any complex numbers.

When b_i = 1 for all i the expression is called a simple continued fraction. When the expression contains finitely many terms, it is called a finite continued fraction. When the expression contains infinitely many terms, it is called an infinite continued fraction.[6] When the terms eventually repeat from some point onwards, the expression is called a periodic continued fraction.[5]

Thus, all of the following illustrate valid finite simple continued fractions:

Examples of finite simple continued fractions

Formula	Numeric	Remarks
${\displaystyle \ a_{0}}$	${\displaystyle \ 2}$	All integers are a degenerate case
${\displaystyle \ a_{0}+{\cfrac {1}{a_{1}}}}$	${\displaystyle \ 2+{\cfrac {1}{3}}}$	Simplest possible fractional form
${\displaystyle \ a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}}}}}}$	${\displaystyle \ -3+{\cfrac {1}{2+{\cfrac {1}{18}}}}}$	First integer may be negative
${\displaystyle \ a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{a_{3}}}}}}}}$	${\displaystyle \ {\cfrac {1}{15+{\cfrac {1}{1+{\cfrac {1}{102}}}}}}}$	First integer may be zero

For simple continued fractions of the form

${\displaystyle r=a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{a_{3}+{_{\ddots }}}}}}}}}$

the ${\displaystyle a_{n}}$ term can be calculated using the following recursive formula:

${\displaystyle a_{n}=\left\lfloor {\frac {N_{n}}{N_{n+1}}}\right\rfloor }$

where ${\displaystyle N_{n+1}=N_{n-1}{\bmod {N}}_{n}}$ and ${\displaystyle {\begin{cases}N_{0}=r\\N_{1}=1\end{cases}}}$

From which it can be understood that the ${\displaystyle a_{n}}$ sequence stops if ${\displaystyle N_{n+1}=0}$.

Consider a real number ${\displaystyle r}$. Let ${\displaystyle i=\lfloor r\rfloor }$ and let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle f = r - i} . When ${\displaystyle f\neq 0}$, the continued fraction representation of ${\displaystyle r}$ is ${\displaystyle [i;a_{1},a_{2},\ldots ]}$, where ${\displaystyle [a_{1};a_{2},\ldots ]}$ is the continued fraction representation of ${\displaystyle 1/f}$. Note: when ${\displaystyle r\geq 0}$, then ${\displaystyle i}$ is the integer part of ${\displaystyle r}$, and ${\displaystyle f}$ is the fractional part of ${\displaystyle r}$.

In order to calculate a continued fraction representation of a number ${\displaystyle r}$, write down the floor of ${\displaystyle r}$. Subtract this value from ${\displaystyle r}$. If the difference is 0, stop; otherwise find the reciprocal of the difference and repeat. The procedure will halt if and only if ${\displaystyle r}$ is rational. This process can be efficiently implemented using the Euclidean algorithm when the number is rational. The table below shows an implementation of this procedure for the number 3.245, resulting in the continued fraction expansion ${\displaystyle [3;4,12,4]}$.

Find the simple continued fraction for ${\displaystyle 3.245={\frac {649}{200}}}$

Step	Real Number	Integer part	Fractional part	Simplified	Reciprocal of f
1	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle r = \frac{649}{200}}	${\displaystyle i=3}$	${\displaystyle f={\frac {649}{200}}-3}$	${\displaystyle ={\frac {49}{200}}}$	${\displaystyle {\frac {1}{f}}={\frac {200}{49}}}$
2	${\displaystyle r={\frac {200}{49}}}$	${\displaystyle i=4}$	${\displaystyle f={\frac {200}{49}}-4}$	${\displaystyle ={\frac {4}{49}}}$	${\displaystyle {\frac {1}{f}}={\frac {49}{4}}}$
3	${\displaystyle r={\frac {49}{4}}}$	${\displaystyle i=12}$	${\displaystyle f={\frac {49}{4}}-12}$	${\displaystyle ={\frac {1}{4}}}$	${\displaystyle {\frac {1}{f}}={\frac {4}{1}}}$
4	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle r = 4}	${\displaystyle i=4}$	${\displaystyle f=4-4}$	${\displaystyle =0}$	STOP
Continued fraction form for ${\displaystyle 3.245={\frac {649}{200}}=[3;4,12,4]}$ = 3 + 1/4 + 1/12 + 1/4

The integers ${\displaystyle a_{0}}$, ${\displaystyle a_{1}}$ etc., are called the coefficients or terms of the continued fraction.[2] One can abbreviate the continued fraction

${\displaystyle x=a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{a_{3}}}}}}}}$

in the notation of Carl Friedrich Gauss

${\displaystyle x=a_{0}+{\underset {i=1}{\overset {3}{\mathrm {K} }}}~{\frac {1}{a_{i}}}}$

or as

${\displaystyle x=[a_{0};a_{1},a_{2},a_{3}]}$,

or in the notation of Pringsheim as

${\displaystyle x=a_{0}+{\frac {1\mid }{\mid a_{1}}}+{\frac {1\mid }{\mid a_{2}}}+{\frac {1\mid }{\mid a_{3}}},}$

or in related notations as

${\displaystyle x=a_{0}+{1 \over a_{1}+{}}{1 \over a_{2}+{}}{1 \over a_{3}{}}}$

or

${\displaystyle x=a_{0}+{1 \over a_{1}}{{} \atop +}{1 \over a_{2}}{{} \atop +}{1 \over a_{3}}.}$

Sometimes angle brackets are used, like this:

${\displaystyle x=\left\langle a_{0};a_{1},a_{2},a_{3}\right\rangle .}$

The semicolon in the square and angle bracket notations is sometimes replaced by a comma.[3][4]

One may also define infinite simple continued fractions as limits:

${\displaystyle [a_{0};a_{1},a_{2},a_{3},\,\ldots ]=\lim _{n\to \infty }[a_{0};a_{1},a_{2},\,\ldots ,a_{n}].}$

This limit exists for any choice of ${\displaystyle a_{0}}$ and positive integers ${\displaystyle a_{1},a_{2},\ldots }$.[7][8]

Every finite continued fraction represents a rational number, and every rational number can be represented in precisely two different ways as a finite continued fraction, with the conditions that the first coefficient is an integer and the other coefficients are positive integers. These two representations agree except in their final terms. In the longer representation the final term in the continued fraction is 1; the shorter representation drops the final 1, but increases the new final term by 1. The final element in the short representation is therefore always greater than 1, if present. In symbols:

[a₀; a₁, a₂, ..., a_{n − 1}, a_n, 1] = [a₀; a₁, a₂, ..., a_{n − 1}, a_n + 1].

[a₀; 1] = [a₀ + 1].

The continued fraction representations of a positive rational number and its reciprocal are identical except for a shift one place left or right depending on whether the number is less than or greater than one respectively. In other words, the numbers represented by ${\displaystyle [a_{0};a_{1},a_{2},\ldots ,a_{n}]}$ and ${\displaystyle [0;a_{0},a_{1},\ldots ,a_{n}]}$ are reciprocals.

For instance if ${\displaystyle a}$ is an integer and ${\displaystyle x<1}$ then

${\displaystyle x=0+{\frac {1}{a+{\frac {1}{b}}}}}$ and ${\displaystyle {\frac {1}{x}}=a+{\frac {1}{b}}}$.

If ${\displaystyle x>1}$ then

${\displaystyle x=a+{\frac {1}{b}}}$ and ${\displaystyle {\frac {1}{x}}=0+{\frac {1}{a+{\frac {1}{b}}}}}$.

The last number that generates the remainder of the continued fraction is the same for both ${\displaystyle x}$ and its reciprocal.

For example,

${\displaystyle 2.25={\frac {9}{4}}=[2;4]}$ and ${\displaystyle {\frac {1}{2.25}}={\frac {4}{9}}=[0;2,4]}$.

Convergents approaching the golden ratio

Every infinite continued fraction is irrational, and every irrational number can be represented in precisely one way as an infinite continued fraction.

An infinite continued fraction representation for an irrational number is useful because its initial segments provide rational approximations to the number. These rational numbers are called the convergents of the continued fraction.[9][10] The larger a term is in the continued fraction, the closer the corresponding convergent is to the irrational number being approximated. Numbers like π have occasional large terms in their continued fraction, which makes them easy to approximate with rational numbers. Other numbers like e have only small terms early in their continued fraction, which makes them more difficult to approximate rationally. The golden ratio Φ has terms equal to 1 everywhere—the smallest values possible—which makes Φ the most difficult number to approximate rationally. In this sense, therefore, it is the "most irrational" of all irrational numbers. Even-numbered convergents are smaller than the original number, while odd-numbered ones are larger.

For a continued fraction [a₀; a₁, a₂, ...], the first four convergents (numbered 0 through 3) are

a₀/1, a₁a₀ + 1/a₁, a₂(a₁a₀ + 1) + a₀/a₂a₁ + 1, a₃(a₂(a₁a₀ + 1) + a₀) + (a₁a₀ + 1)/ a₃(a₂a₁ + 1) + a₁.

The numerator of the third convergent is formed by multiplying the numerator of the second convergent by the third coefficient, and adding the numerator of the first convergent. The denominators are formed similarly. Therefore, each convergent can be expressed explicitly in terms of the continued fraction as the ratio of certain multivariate polynomials called continuants.

If successive convergents are found, with numerators h₁, h₂, ... and denominators k₁, k₂, ... then the relevant recursive relation is that of Gaussian brackets:

h_n = a_nh_{n − 1} + h_{n − 2},

k_n = a_nk_{n − 1} + k_{n − 2}.

The successive convergents are given by the formula

h_n/k_n = a_nh_{n − 1} + h_{n − 2}/a_nk_{n − 1} + k_{n − 2}.

Thus to incorporate a new term into a rational approximation, only the two previous convergents are necessary. The initial "convergents" (required for the first two terms) are ⁰⁄₁ and ¹⁄₀. For example, here are the convergents for [0;1,5,2,2].

n	−2	−1	0	1	2	3	4
an			0	1	5	2	2
hn	0	1	0	1	5	11	27
kn	1	0	1	1	6	13	32

When using the Babylonian method to generate successive approximations to the square root of an integer, if one starts with the lowest integer as first approximant, the rationals generated all appear in the list of convergents for the continued fraction. Specifically, the approximants will appear on the convergents list in positions 0, 1, 3, 7, 15, ... , 2^k−1, ... For example, the continued fraction expansion for √3 is [1;1,2,1,2,1,2,1,2,...]. Comparing the convergents with the approximants derived from the Babylonian method:

n	−2	−1	0	1	2	3	4	5	6	7
an			1	1	2	1	2	1	2	1
hn	0	1	1	2	5	7	19	26	71	97
kn	1	0	1	1	3	4	11	15	41	56

x₀ = 1 = 1/1

x₁ = 1/2(1 + 3/1) = 2/1 = 2

x₂ = 1/2(2 + 3/2) = 7/4

x₃ = 1/2(7/4 + 3/7/4) = 97/56

Some useful theorems

If ${\displaystyle \ a_{0}\ ,}$ ${\displaystyle a_{1}\ ,}$ ${\displaystyle a_{2}\ ,}$ ${\displaystyle \ \ldots \ }$ is an infinite sequence of positive integers, define the sequences ${\displaystyle \ h_{n}\ }$ and ${\displaystyle \ k_{n}\ }$ recursively:

${\displaystyle h_{n}=a_{n}\ h_{n-1}+h_{n-2}\ ,}$			${\displaystyle h_{-1}=1\ ,}$		${\displaystyle h_{-2}=0\$ ;}
${\displaystyle k_{n}=a_{n}\ k_{n-1}+k_{n-2}\ ,}$			${\displaystyle k_{-1}=0\ ,}$		${\displaystyle k_{-2}=1~.}$

Theorem 1. For any positive real number ${\displaystyle \ x\ }$

${\displaystyle \left[\ a_{0};\ a_{1},\ \dots ,a_{n-1},x\ \right]={\frac {x\ h_{n-1}+h_{n-2}}{\ x\ k_{n-1}+k_{n-2}\ }}~.}$

Theorem 2. The convergents of ${\displaystyle \ [\ a_{0}\$ ;} ${\displaystyle a_{1}\ ,}$ ${\displaystyle a_{2}\ ,}$ ${\displaystyle \ldots \ ]\ }$ are given by

${\displaystyle \left[\ a_{0};\ a_{1},\ \dots ,a_{n}\ \right]={\frac {h_{n}}{\ k_{n}\ }}~.}$

Theorem 3. If the ${\displaystyle \ n}$th convergent to a continued fraction is ${\displaystyle \ {\frac {h_{n}}{k_{n}}}\ ,}$ then

${\displaystyle k_{n}\ h_{n-1}-k_{n-1}\ h_{n}=(-1)^{n}\ ,}$

or equivalently

${\displaystyle {\frac {h_{n}}{\ k_{n}\ }}-{\frac {h_{n-1}}{\ k_{n-1}\ }}={\frac {(-1)^{n+1}}{\ k_{n-1}\ k_{n}\ }}~.}$

Corollary 1: Each convergent is in its lowest terms (for if ${\displaystyle \ h_{n}\ }$ and ${\displaystyle \ k_{n}\ }$ had a nontrivial common divisor it would divide ${\displaystyle \ k_{n}\ h_{n-1}-k_{n-1}\ h_{n}\ ,}$ which is impossible).

Corollary 2: The difference between successive convergents is a fraction whose numerator is unity:

${\displaystyle {\frac {h_{n}}{k_{n}}}-{\frac {h_{n-1}}{k_{n-1}}}={\frac {\ h_{n}\ k_{n-1}-k_{n}\ h_{n-1}\ }{\ k_{n}\ k_{n-1}\ }}={\frac {(-1)^{n+1}}{\ k_{n}\ k_{n-1}\ }}~.}$

Corollary 3: The continued fraction is equivalent to a series of alternating terms:

${\displaystyle a_{0}+\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{\ k_{n}\ k_{n+1}\ }}~.}$

Corollary 4: The matrix

${\displaystyle {\begin{bmatrix}h_{n}&h_{n-1}\\k_{n}&k_{n-1}\end{bmatrix}}}$

has determinant plus or minus one, and thus belongs to the group of ${\displaystyle \ 2\times 2\ }$ unimodular matrices ${\displaystyle \ \mathrm {GL} (2,\mathbb {Z} )~.}$

Theorem 4. Each (${\displaystyle \ s}$th) convergent is nearer to a subsequent (${\displaystyle \ n}$th) convergent than any preceding (${\displaystyle \ r}$th) convergent is. In symbols, if the ${\displaystyle \ n}$th convergent is taken to be ${\displaystyle \ \left[\ a_{0};\ a_{1},\ \ldots ,\ a_{n}\ \right]=x_{n}\ ,}$ then

${\displaystyle \left|\ x_{r}-x_{n}\ \right|>\left|\ x_{s}-x_{n}\ \right|}$

for all ${\displaystyle \ r<s<n~.}$

Corollary 1: The even convergents (before the ${\displaystyle \ n}$th) continually increase, but are always less than ${\displaystyle \ x_{n}~.}$

Corollary 2: The odd convergents (before the ${\displaystyle \ n}$th) continually decrease, but are always greater than ${\displaystyle \ x_{n}~.}$

Theorem 5.

${\displaystyle {\frac {1}{\ k_{n}\ (k_{n+1}+k_{n})\ }}<\left|\ x-{\frac {h_{n}}{\ k_{n}\ }}\ \right|<{\frac {1}{\ k_{n}\ k_{n+1}\ }}~.}$

Corollary 1: A convergent is nearer to the limit of the continued fraction than any fraction whose denominator is less than that of the convergent.

Corollary 2: A convergent obtained by terminating the continued fraction just before a large term is a close approximation to the limit of the continued fraction.

If

${\displaystyle {\frac {h_{n-1}}{k_{n-1}}},{\frac {h_{n}}{k_{n}}}}$

are consecutive convergents, then any fractions of the form

${\displaystyle {\frac {h_{n-1}+mh_{n}}{k_{n-1}+mk_{n}}},}$

where ${\displaystyle m}$ is an integer such that ${\displaystyle 0\leq m\leq a_{n+1}}$, are called semiconvergents, secondary convergents, or intermediate fractions. The ${\displaystyle (m+1)}$-st semiconvergent equals the mediant of the ${\displaystyle m}$-th one and the convergent ${\displaystyle {\tfrac {h_{n}}{k_{n}}}}$. Sometimes the term is taken to mean that being a semiconvergent excludes the possibility of being a convergent (i.e., ${\displaystyle 0<m<a_{n+1}}$), rather than that a convergent is a kind of semiconvergent.

It follows that semiconvergents represent a monotonic sequence of fractions between the convergents ${\displaystyle {\tfrac {h_{n-1}}{k_{n-1}}}}$ (corresponding to ${\displaystyle m=0}$) and ${\displaystyle {\tfrac {h_{n+1}}{k_{n+1}}}}$ (corresponding to ${\displaystyle m=a_{n+1}}$). The consecutive semiconvergents ${\displaystyle {\tfrac {a}{b}}}$ and ${\displaystyle {\tfrac {c}{d}}}$ satisfy the property ${\displaystyle ad-bc=\pm 1}$.

If a rational approximation ${\displaystyle {\tfrac {p}{q}}}$ to a real number ${\displaystyle x}$ is such that the value ${\displaystyle \left|x-{\tfrac {p}{q}}\right|}$ is smaller than that of any approximation with a smaller denominator, then ${\displaystyle {\tfrac {p}{q}}}$ is a semiconvergent of the continued fraction expansion of ${\displaystyle x}$. The converse is not true, however.

One can choose to define a best rational approximation to a real number x as a rational number n/d, d > 0, that is closer to x than any approximation with a smaller or equal denominator. The simple continued fraction for x can be used to generate all of the best rational approximations for x by applying these three rules:

1. Truncate the continued fraction, and reduce its last term by a chosen amount (possibly zero).
2. The reduced term cannot have less than half its original value.
3. If the final term is even, half its value is admissible only if the corresponding semiconvergent is better than the previous convergent. (See below.)

For example, 0.84375 has continued fraction [0;1,5,2,2]. Here are all of its best rational approximations.

Continued fraction	[0;1]	[0;1,3]	[0;1,4]	[0;1,5]	[0;1,5,2]	[0;1,5,2,1]	[0;1,5,2,2]
Rational approximation	1	3/4	4/5	5/6	11/13	16/19	27/32
Decimal equivalent	1	0.75	0.8	~0.83333	~0.84615	~0.84211	0.84375
Error	+18.519%	−11.111%	−5.1852%	−1.2346%	+0.28490%	−0.19493%	0%

Best rational approximants for π (green circle), e (blue diamond), ϕ (pink oblong), (√3)/2 (grey hexagon), 1/√2 (red octagon) and 1/√3 (orange triangle) calculated from their continued fraction expansions, plotted as slopes y/x with errors from their true values (black dashes)  

The strictly monotonic increase in the denominators as additional terms are included permits an algorithm to impose a limit, either on size of denominator or closeness of approximation.

The "half rule" mentioned above requires that when a_k is even, the halved term a_k/2 is admissible if and only if |x − [a₀ ; a₁, ..., a_{k − 1}]| > |x − [a₀ ; a₁, ..., a_{k − 1}, a_k/2]| [11] This is equivalent [11] to: Shoemake (1995).

[a_k; a_{k − 1}, ..., a₁] > [a_k; a_{k + 1}, ...].

The convergents to x are "best approximations" in a much stronger sense than the one defined above. Namely, n/d is a convergent for x if and only if |dx − n| has the smallest value among the analogous expressions for all rational approximations m/c with c ≤ d; that is, we have |dx − n| < |cx − m| so long as c < d. (Note also that |d_kx − n_k| → 0 as k → ∞.)

Consider x = [a₀; a₁, ...] and y = [b₀; b₁, ...]. If k is the smallest index for which a_k is unequal to b_k then x < y if (−1)^k(a_k − b_k) < 0 and y < x otherwise.

If there is no such k, but one expansion is shorter than the other, say x = [a₀; a₁, ..., a_n] and y = [b₀; b₁, ..., b_n, b_{n + 1}, ...] with a_i = b_i for 0 ≤ i ≤ n, then x < y if n is even and y < x if n is odd.

To calculate the convergents of π we may set a₀ = ⌊π⌋ = 3, define u₁ = 1/π − 3 ≈ 7.0625 and a₁ = ⌊u₁⌋ = 7, u₂ = 1/u₁ − 7 ≈ 15.9966 and a₂ = ⌊u₂⌋ = 15, u₃ = 1/u₂ − 15 ≈ 1.0034. Continuing like this, one can determine the infinite continued fraction of π as

[3;7,15,1,292,1,1,...] (sequence A001203 in the OEIS).

The fourth convergent of π is [3;7,15,1] = 355/113 = 3.14159292035..., sometimes called Milü, which is fairly close to the true value of π.

Let us suppose that the quotients found are, as above, [3;7,15,1]. The following is a rule by which we can write down at once the convergent fractions which result from these quotients without developing the continued fraction.

The first quotient, supposed divided by unity, will give the first fraction, which will be too small, namely, 3/1. Then, multiplying the numerator and denominator of this fraction by the second quotient and adding unity to the numerator, we shall have the second fraction, 22/7, which will be too large. Multiplying in like manner the numerator and denominator of this fraction by the third quotient, and adding to the numerator the numerator of the preceding fraction, and to the denominator the denominator of the preceding fraction, we shall have the third fraction, which will be too small. Thus, the third quotient being 15, we have for our numerator (22 × 15 = 330) + 3 = 333, and for our denominator, (7 × 15 = 105) + 1 = 106. The third convergent, therefore, is 333/106. We proceed in the same manner for the fourth convergent. The fourth quotient being 1, we say 333 times 1 is 333, and this plus 22, the numerator of the fraction preceding, is 355; similarly, 106 times 1 is 106, and this plus 7 is 113. In this manner, by employing the four quotients [3;7,15,1], we obtain the four fractions:

3/1, 22/7, 333/106, 355/113, ....

The following Maple code will generate continued fraction expansions of pi

To sum up, the pattern is ${\displaystyle {\text{Numerator}}_{i}={\text{Numerator}}_{(i-1)}\cdot {\text{Quotient}}_{i}+{\text{Numerator}}_{(i-2)}}$ ${\displaystyle {\text{Denominator}}_{i}={\text{Denominator}}_{(i-1)}\cdot {\text{Quotient}}_{i}+{\text{Denominator}}_{(i-2)}}$

These convergents are alternately smaller and larger than the true value of π, and approach nearer and nearer to π. The difference between a given convergent and π is less than the reciprocal of the product of the denominators of that convergent and the next convergent. For example, the fraction 22/7 is greater than π, but 22/7 − π is less than 1/7 × 106 = 1/742 (in fact, 22/7 − π is just more than 1/791 = 1/7 × 113).

The demonstration of the foregoing properties is deduced from the fact that if we seek the difference between one of the convergent fractions and the next adjacent to it we shall obtain a fraction of which the numerator is always unity and the denominator the product of the two denominators. Thus the difference between 22/7 and 3/1 is 1/7, in excess; between 333/106 and 22/7, 1/742, in deficit; between 355/113 and 333/106, 1/11978, in excess; and so on. The result being, that by employing this series of differences we can express in another and very simple manner the fractions with which we are here concerned, by means of a second series of fractions of which the numerators are all unity and the denominators successively be the product of every two adjacent denominators. Instead of the fractions written above, we have thus the series:

3/1 + 1/1 × 7 − 1/7 × 106 + 1/106 × 113 − ...

The first term, as we see, is the first fraction; the first and second together give the second fraction, 22/7; the first, the second and the third give the third fraction 333/106, and so on with the rest; the result being that the series entire is equivalent to the original value.

A generalized continued fraction is an expression of the form

${\displaystyle x=b_{0}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+{\cfrac {a_{3}}{b_{3}+{\cfrac {a_{4}}{b_{4}+\ddots \,}}}}}}}}}$

where the a_n (n > 0) are the partial numerators, the b_n are the partial denominators, and the leading term b₀ is called the integer part of the continued fraction.

To illustrate the use of generalized continued fractions, consider the following example. The sequence of partial denominators of the simple continued fraction of π does not show any obvious pattern:

${\displaystyle \pi =[3;7,15,1,292,1,1,1,2,1,3,1,\ldots ]}$

or

${\displaystyle \pi =3+{\cfrac {1}{7+{\cfrac {1}{15+{\cfrac {1}{1+{\cfrac {1}{292+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{3+{\cfrac {1}{1+\ddots }}}}}}}}}}}}}}}}}}}}}}}$

However, several generalized continued fractions for π have a perfectly regular structure, such as:

${\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+{\cfrac {9^{2}}{2+\ddots }}}}}}}}}}}}={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}=3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+{\cfrac {7^{2}}{6+{\cfrac {9^{2}}{6+\ddots }}}}}}}}}}}$

${\displaystyle \displaystyle \pi =2+{\cfrac {2}{1+{\cfrac {1}{1/2+{\cfrac {1}{1/3+{\cfrac {1}{1/4+\ddots }}}}}}}}=2+{\cfrac {2}{1+{\cfrac {1\cdot 2}{1+{\cfrac {2\cdot 3}{1+{\cfrac {3\cdot 4}{1+\ddots }}}}}}}}}$

${\displaystyle \displaystyle \pi =2+{\cfrac {4}{3+{\cfrac {1\cdot 3}{4+{\cfrac {3\cdot 5}{4+{\cfrac {5\cdot 7}{4+\ddots }}}}}}}}}$

The first two of these are special cases of the arctangent function with π = 4 arctan (1) and the fourth and fifth one can be derived using the Wallis product.[12][13]

${\displaystyle \pi =3+{\cfrac {1}{6+{\cfrac {1^{3}+2^{3}}{6\cdot 1^{2}+1^{2}{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}}{6\cdot 2^{2}+2^{2}{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}}{6\cdot 3^{2}+3^{2}{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}+7^{3}+8^{3}}{6\cdot 4^{2}+\ddots }}}}}}}}}}}$

The continued fraction of ${\displaystyle \pi }$ above consisting of cubes uses the Nilakantha series and an exploit from Leonhard Euler.[14]