I do not believe anything very certainly,
but everything very probably.
​Christian Huygens
Euler proved in his E123 paper, written 1739, a generalizations of Brouncker’s and Wallis’s formulas established in the Arithmetica Infinitorum published 1656.
SOURCES:
​https://download.uni-mainz.de/mathematik/Algebraische%20Geometrie/Euler-Kreis%20Mainz/E123RevisedVersion1.pdf​
​https://scholarlycommons.pacific.edu/euler-works/123/​
Written 1739 Published 1750.
https://mathshistory.st-andrews.ac.uk/Biographies/Brouncker/​​Brouncker’s mathematical achievements includes work on continued fractions and calculating logarithms by infinite series. In 1655 he gave a continued fraction expansion of
□=
4
π
=1+
2
1
2+
2
3
2+
2
5
2+...
​This result, written up in around ten pages, was added by Wallis to his treatise Arithmetica Infinitorum and probably first discovered by Brouncker in 1654. Wallis told Huygens of this result and
Huygens
expressed strong doubts that it was true.
However after Brouncker correctly computed the first 10 places in the decimal expansion of π using his continued fraction expansion, Huygens accepted the result.
Wallis presented in his 1656 Arithmetica Infinitorum an infinite product:
□=
4
π
=
3·3·5·5·7·7·9·etc.
2·4·4·6·6·8·8·etc.
​And Brouncker found an infinite continued fraction:
□=
4
π
=1+
1
2+
9
2+
25
2+
49
2+etc.
​Neither Brouncker nor Wallis gave a proof for the continued fraction formula.
Wallis used the notation □, for which
4
π
is written today.
First we check Wallis infinite product with ContinuedFractionK, Inactive and Activate:
Out[]=
4
π

3·3
2·4
5·5
4·6
7·7
6·8
9·9
8·10
11·11
10·12
etc.
etc.
And then we check Brouncker’s infinite continued fraction:
Out[]=
4
π
1+
1
2+
9
2+
25
2+
49
2+etc.
Now there are other beautiful such infinite continued fractions like:
Out[]=
π3+
1
6+
9
6+
25
6+
49
6+etc.
In E123 Euler was looking for a general solution of this continued fraction:
a-1+
1
(2a-1)+
9
(2a-1)+
25
(2a-1)+
49
(2a-1)+etc.
Euler gives the following solution in E123 §16:(1)
J(α)=
a
1
∫
0
a+1
x
1-
4
x
x
1
∫
0
a-1
x
1-
4
x
x
=
aa(a+4)(a+4)(a+8)etc.
(a+2)(a+2)(a+6)(a+6)etc.
=a-1+
1
(2a-1)+
9
(2a-1)+
25
(2a-1)+
49
(2a-1)+etc.
​​https://download.uni-mainz.de/mathematik/Algebraische%20Geometrie/Euler-Kreis%20Mainz/E123RevisedVersion1.pdf​​i.e. an infinite product and an infinite continued fraction.
First we calculate the recurrence relation of (1) “the hard way”.Step 1: Compute the derivative of
α
x
1-
4
x
:
Out[]=
∂
x

α
x
1-
4
x

-1+α
x
α
1-
4
x
-
3+α
x
(2+α)
1-
4
x
Step 2: Compute the antiderivative:
Out[]=
1
∫
0
∂
x

α
x
1-
4
x
x
α
x
1-
4
x
1
|
0
0
1
∫
0
-1+α
x
α
1-
4
x
-
3+α
x
(2+α)
1-
4
x
x
The recurrence relation is:
1
∫
0
α+3
x
1-
4
x
x=
α
α+2
1
∫
0
α-1
x
1-
4
x
x
​i.e.
1
∫
0
α+1
x
1-
4
x
x=
α-2
α
1
∫
0
α-3
x
1-
4
x
x
and
1
∫
0
α-1
x
1-
4
x
x=
α-4
α-2
1
∫
0
α-5
x
1-
4
x
x
Inserting this into (1) gives: ​
J(α)=
a
1
∫
0
a+1
x
1-
4
x
x
1
∫
0
a-1
x
1-
4
x
x
=
α
α-2
α
α-4
α-2
α-4
α-4
1
∫
0
a-3
x
1-
4
x
1
∫
0
a-5
x
1-
4
x
=
2
(α-2)
2
(α-4)
J(α-4)
So α decreases by 4 at each computation step. You have to compute
J(1),J(2),J(3)
and
J(4)
to stop the recurrence starting with a natural number. But it is also possible to compute
J(1/2),J(3/2),J(5/2)
and J(7/2):
In[]:=
Remove[EulerWallisBrouncker]
In[]:=
EulerWallisBrouncker[n_]/;0<n<=2:=EulerWallisBrouncker[n]=inert
2
2Gamma[inert[2n+4]/8]
Gamma[inert[2n]/8]
​​EulerWallisBrouncker[n_]/;2<n<=4:=EulerWallisBrouncker[n]=inert
2
(4n-8)Gamma[inert[2n-4]/8]
8Gamma[inert[2n]/8]
​​EulerWallisBrouncker[n_]/;n>4:=inert
2
inert[Numerator[n-2]]
2
inert[Numerator[n-4]]
EulerWallisBrouncker[n-4]​​(*Thisistherecurrencerelationof(1)*)
Example: EulerWallisBrouncker[2]
In[]:=
TraditionalForm[EulerWallisBrouncker[2]]
Out[]//TraditionalForm=
4
2
Γ
8
8

2
Γ
4
8

This can be simplified:
In[]:=
%/.inert->Identity
Out[]=
4
π
And if you start with a bigger number you see the recurrence at work, EulerWallisBrouncker[2+8]:
In[]:=
TraditionalForm[EulerWallisBrouncker[2+8]]
Out[]//TraditionalForm=
2
4
2
2
2
8
2
6
4
2
Γ
8
8

2
Γ
4
8

And simplified:
In[]:=
%/.inert->Identity
Out[]=
256
9π
Now the continued fraction is insertes on the right site:
In[]:=
EulerBrounckerContinuedFraction[a_,n_:5]:=TraditionalForm@EulerWallisBrouncker[a]==(a-1)+ContinuedFractionK[inert[
2
(2k-1)
],inert[2(a-1)],{k,1,n}]/.inert[
2
(2n-1)
]inert[2(a-1)]->inert["etc."]
This is Brouncker’s continued fraction:
In[]:=
EulerBrounckerContinuedFraction[2]
Out[]=
4
2
Γ
8
8

2
Γ
4
8

1+
1
2+
9
2+
25
2+
49
2+etc.
And the left side simplifies to:
In[]:=
%/.inert->Identity
Out[]=
4
π
1+
1
2+
9
2+
25
2+
49
2+etc.
In[]:=
EulerBrounckerContinuedFraction[2+8]
Out[]=
2
4
2
2
2
8
2
6
4
2
Γ
8
8

2
Γ
4
8

9+
1
18+
9
18+
25
18+
49
18+etc.
This example was also given by Brouncker/Wallis:
Here are the first 16 Brouncker continued fractions:
Needless to say, that this is also possible:
This can be assigned to a function BrounckerEuler[a]:
ComplexPlot of BrounckerEuler exhibits its beauty:
Now we can check the recurrence relation of (1) directly:
BrounckerEuler[a] is more compact than the code used in EulerBrounckerContinuedFraction[a], but gives the same answer:
cf with BrounckerEuler:
Here are the generalized Wallis products:
Check with Mathematica directly:
Here are another version of the generalized Wallis products:
And yes, this is also possible:

Further Reading

The Arithmetic of Infinitesimals, John Wallis 1656
Translated from Latin to English with an Introduction by
Jacqueline A. Stedall.
Arithmetica Infinitorum, John Wallis 1656
​https://archive.org/details/ArithmeticaInfinitorum/page/n209/mode/2up
These are the continued fractions of Brouncker for example I:

Appendix: Euler/Ramanujan’s Continued Fraction Formula

Source: Bruce Berndt, Ramanujan’s Notebooks Part 2 (1989), Chapter 12 Continued Fractions, p.145:
Taking the reciprocals on both sides times 4 is even simpler:
Example: Continued fraction
Taking 4 times the reciprocals:
Some stranger continued fraction:
Cf to EulerBrounckerContinuedFraction,the right hand side are the same (the parameter is shifted by 1):
And just to be sure, the values are the same on the left hand side:

Addendum: Lambert’s Continued Fraction for π/4

https://hal.science/hal-02984214/document​
Denis Roegel. Lambert’s proof of the irrationality of Pi: Context and translation. [Research Report] LORIA. 2020. hal-02984214
The well known arctan series:
Can be converted into a continued fraction (Pringsheim notation):
The continued fraction can be turned back into the arctan series:
This is even nicer:

Wolfram Mathematica Code

https://en.wikipedia.org/wiki/Alfred_Pringsheim​
​https://tex.stackexchange.com/questions/68190/how-do-i-typeset-continued-fractions-in-pringsheims-concise-notation

CITE THIS NOTEBOOK

Brouncker's continued fraction of Wallis's Arithmetica Infinitorum and Euler E123​
by Oliver Seipel​
Wolfram Community, STAFF PICKS, April 22, 2024
​https://community.wolfram.com/groups/-/m/t/3162556