What are convergents and why are they useful?

Convergents are the rational numbers obtained by truncating a continued fraction after each term. They are the best rational approximations to the original number in a precise mathematical sense: no fraction with a smaller denominator is closer to the target. The fraction 355/113 (accurate to 7 digits) is a convergent of π.

Why does 355/113 approximate π so well?

The continued fraction for π is [3; 7, 15, 1, 292, …]. Stopping after [3; 7, 15, 1] gives the convergent 355/113 ≈ 3.14159292…, which agrees with π to 6 decimal places. The large next coefficient (292) indicates this convergent is an unusually good approximation.

What is the difference between finite and infinite continued fractions?

Rational numbers always yield finite continued fractions. For example, 355/113 = [3; 7, 16]. Irrational numbers yield infinite continued fractions. This tool shows up to 20 terms for irrational inputs.

How accurate are the convergents?

Each convergent pₙ/qₙ satisfies |x − pₙ/qₙ| < 1/(qₙ·qₙ₊₁). The larger the next coefficient aₙ₊₁, the better the current convergent approximates x. This makes continued fractions the optimal tool for finding rational approximations.

Continued Fraction Converter - Best Rational Approximations

#	CF terms used	Fraction	Decimal	Error
1	[3]	3/1	3.000000000	1.42e-1
2	[3; 7]	22/7	3.142857143	1.26e-3
3	[3; 7, 15]	333/106	3.141509434	8.32e-5
4	[3; 7, 15, 1]	355/113	3.141592920	2.70e-7
5	[3; 7, 15, 1, 288]	102573/32650	3.141592649	6.89e-10
6	[3; 7, 15, 1, 288, 1]	102928/32763	3.141592650	2.46e-10
7	[3; 7, 15, 1, 288, 1, 2]	308429/98176	3.141592650	6.52e-11
8	[3; 7, 15, 1, 288, 1, 2, 1]	411357/130939	3.141592650	1.26e-11
9	[3; 7, 15, 1, 288, 1, 2, 1, 3]	1542500/490993	3.141592650	2.95e-12
10	[3; 7, 15, 1, 288, 1, 2, 1, 3, 1]	1953857/621932	3.141592650	3.22e-13
11	[3; 7, 15, 1, 288, 1, 2, 1, 3, 1, 7]	15219499/4844517	3.141592650	1.07e-14
12	[3; 7, 15, 1, 288, 1, 2, 1, 3, 1, 7, 4]	62831853/20000000	3.141592650	0

CF terms used

Fraction

Decimal

Error

[3]

3/1

3.000000000

1.42e-1

[3; 7]

22/7

3.142857143

1.26e-3

[3; 7, 15]

333/106

3.141509434

8.32e-5

[3; 7, 15, 1]

355/113

3.141592920

2.70e-7

[3; 7, 15, 1, 288]

102573/32650

3.141592649

6.89e-10

[3; 7, 15, 1, 288, 1]

102928/32763

3.141592650

2.46e-10

[3; 7, 15, 1, 288, 1, 2]

308429/98176

3.141592650

6.52e-11

[3; 7, 15, 1, 288, 1, 2, 1]

411357/130939

3.141592650

1.26e-11

[3; 7, 15, 1, 288, 1, 2, 1, 3]

1542500/490993

3.141592650

2.95e-12

[3; 7, 15, 1, 288, 1, 2, 1, 3, 1]

1953857/621932

3.141592650

3.22e-13

[3; 7, 15, 1, 288, 1, 2, 1, 3, 1, 7]

15219499/4844517

3.141592650

1.07e-14

[3; 7, 15, 1, 288, 1, 2, 1, 3, 1, 7, 4]

62831853/20000000

3.141592650

What is a continued fraction?

A continued fraction is an expression of the form a₀ + 1/(a₁ + 1/(a₂ + 1/(a₃ + …))), written compactly as [a₀; a₁, a₂, …]. Every real number has a continued fraction expansion, and every rational number has a finite one.

Famous continued fractions

π = [3; 7, 15, 1, 292, 1, 1, 1, 2, …]
e = [2; 1, 2, 1, 1, 4, 1, 1, 6, …] (beautiful pattern!)
√2 = [1; 2, 2, 2, 2, …] (periodic)
φ (golden ratio) = [1; 1, 1, 1, …] (all ones: "worst" approximable number)
355/113 = [3; 7, 16]: an unusually precise approximation of π

Why are convergents the best approximations?

A convergent pₙ/qₙ is closer to the target than any other fraction with denominator ≤ qₙ. This makes continued fractions the theoretically optimal tool for rational approximation: it's important in gear design, music theory (equal temperament), and calendar design (when is a year divisible by a simple fraction?).

The significance of large coefficients

A large coefficient aₖ (like 292 in π's expansion) means the previous convergent is an exceptionally good approximation. The fraction 355/113 is so accurate because the next coefficient is 292, meaning the next improvement requires a denominator roughly 292 times larger.

Connection to the Euclidean algorithm

The continued fraction expansion of a rational number a/b (where a > b) produces exactly the same sequence of quotients as running the Euclidean GCD algorithm on a and b. For example, to find gcd(34, 21):

34 = 1×21 + 13  -> quotient 1
21 = 1×13 +  8  -> quotient 1
13 = 1× 8 +  5  -> quotient 1
 8 = 1× 5 +  3  -> quotient 1
 5 = 1× 3 +  2  -> quotient 1
 3 = 1× 2 +  1  -> quotient 1
 2 = 2× 1 +  0  -> quotient 2

So 34/21 = [1; 1, 1, 1, 1, 1, 2] in continued fraction notation - the Fibonacci quotients, which is why consecutive Fibonacci numbers are the "worst case" for the Euclidean algorithm (requiring the most steps for their size). This deep connection between continued fractions and the GCD algorithm was known to Euler.