Life of π: Continued Fractions and Infinite Series

Outlines

Life of π: Continued Fractions and Inﬁnite Series

Daniel J. Hermes

February 29, 2012

Daniel J. Hermes Life of π: Continued Fractions and Inﬁnite Series

Outlines

1
π =3+
9
6+
25
6+
49
6+
.
6 + ..


Part I: Introductory Facts
Outlines
Part II: What We Came Here For: π

Outline of Part I
Introduction
The Wrong Way
Smart Men
Conway and Pell


Outlines

Outline of Part I
Introduction
The Wrong Way
Smart Men
Conway and Pell
Continued Fractions
Deﬁne It
What Does it Mean to Converge?
General


Outlines

Outline of Part I
Introduction
The Wrong Way
Smart Men
Conway and Pell
Continued Fractions
Deﬁne It
General
Working with the Convergents
Diﬀerence
Telescoping Partials


Outlines

Outline of Part I
Introduction
The Wrong Way
Smart Men
Conway and Pell
Continued Fractions
Deﬁne It
General
Diﬀerence
Lemmata: Sums for π
Review
Smith Sum

Outlines

Outline of Part II

Analysis and Arithmetic
A Nifty Identity
Denominator Series (kn )
Convergence
Plug and Play
Plug and Play


Outlines

Outline of Part II

A Nifty Identity
Convergence
Plug and Play
Plug and Play

Parting Words
4 12
π and π 2
Thank You


Intro
CF
Convergents

Part I

Introductory Facts


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

Introduction
The Wrong Way
Smart Men
Conway and Pell
Continued Fractions
Deﬁne It
General
Diﬀerence
Review
Smith Sum


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

The Wrong Way
In bill #246 of the 1897 sitting of the Indiana General Assembly, π
was rational:

http://en.wikipedia.org/wiki/Indiana Pi Bill


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

The Wrong Way
was rational:
...Furthermore, it has revealed the ratio of the chord and
arc of ninety degrees, which is as seven to eight, and also
the ratio of the diagonal and one side of a square which
is as ten to seven, disclosing the fourth important fact,
that the ratio of the diameter and circumference is as
ﬁve-fourths to four...



Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

The Wrong Way
was rational:
...Furthermore, it has revealed the ratio of the chord and
arc of ninety degrees, which is as seven to eight, and also
the ratio of the diagonal and one side of a square which
is as ten to seven, disclosing the fourth important fact,
that the ratio of the diameter and circumference is as
ﬁve-fourths to four...

4
π= 5/4
= 3.2



Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

Smart Men: Archimedes

In the third century BCE, Archimedes proved the sharp
inequalities
223/71 < π < 22/7

by means of regular 96-gons


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

Smart Men: Archimedes

In the third century BCE, Archimedes proved the sharp
inequalities
223/71 < π < 22/7

by means of regular 96-gons

22 1
=3+
7 7


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

Smart Men: The Bible
In 1 Kings 7:23 the word translated ’measuring line’
appears in the Hebrew text ... The ratio of the numerical
values of these Hebrew spellings is 111/106. If the putative
value of 3 is multiplied by this ratio, one obtains 333/106


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell


333 15
=3+
106 106


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell


333 15
=3+
106 106
1
= 3 + 106
/15


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell


333 15
=3+
106 106
1
= 3 + 106
/15
1
=3+
1
7+
15

Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

Smart Men: Zu Chongzhi
The 5th century Chinese mathematician and astronomer
Zu Chongzhi ... gave two other approximations ... 22/7
and 355/113


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

and 355/113

355 16
=3+
113 113


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

and 355/113

355 16
=3+
113 113
1 1
= 3 + 113 = 3 +
/16 7 + 1/16


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

and 355/113

355 16
=3+
113 113
1 1
= 3 + 113 = 3 +
/16 7 + 1/16
1
=3+
1
7+
15 + 1/1


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

Smart Men: James Gregory

In 1672, James Gregory wrote about a formula for calculating the
angle given the tangent x:


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell


In 1672, James Gregory wrote about a formula for calculating the
angle given the tangent x:

x3 x5 x7
arctan x = x − + − + ···
3 5 7


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

Why is this relevant? We can use it to approximate π!


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

∞
π 1 1 1 (−1)n
= 1 − + − + ··· =
4 3 5 7 2n + 1
n=0


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

∞
π 1 1 1 (−1)n
= 1 − + − + ··· =
4 3 5 7 2n + 1
n=0

1
Proof: Since d
dx arctan x = = 1 − x 2 + x 4 − x 6 + · · · and
1 + x2
arctan 0 = 0, integrating, we ﬁnd the Taylor series expansion for
arctan is


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

∞
π 1 1 1 (−1)n
= 1 − + − + ··· =
4 3 5 7 2n + 1
n=0

1
Proof: Since d
1 + x2
arctan is
x3 x5 x7
arctan x = x − + − + ···
3 5 7


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

∞
π 1 1 1 (−1)n
= 1 − + − + ··· =
4 3 5 7 2n + 1
n=0

1
Proof: Since d
1 + x2
arctan is
x3 x5 x7
arctan x = x − + − + ···
3 5 7
∞
π (−1)n
Hence = arctan(1) = .
4 2n + 1
n=0


Intro
The Wrong Way
CF
Smart Men
Convergents
Conway and Pell

Conway and Pell

In the actual talk, I displayed “Hey Dan, go to the chalkboard
please.” and just talked to them. For some reference see:
blog.bossylobster.com/2011/07/continued-fractions-for-greater-good.html

blog.bossylobster.com/2011/07/continued-fraction-expansions-of.html

blog.bossylobster.com/2011/08/conways-topograph-part-1.html



blog.bossylobster.com/2011/08/finding-fibonacci-golden-nuggets.html

blog.bossylobster.com/2011/08/finding-fibonacci-golden-nuggets-part-2.html


Intro
Define It
CF
Convergents
General

Introduction
The Wrong Way
Smart Men
Conway and Pell
Continued Fractions
Define It
General
Difference
Review
Smith Sum


Intro
Define It
CF
Convergents
General

Define It

Define a standard continued fraction with a series of nonnegative
integers {an }∞ :
n=0


Intro
Define It
CF
Convergents
General

Define It

Define a standard continued fraction with a series of nonnegative
integers {an }∞ :
n=0

1
a0 +
1
a1 +
1
a2 +
.
a3 + . .


Intro
Deﬁne It
CF
Convergents
General

Deﬁne It

How will we know if our fractions converge if we don’t have a
concept of a “convergent sum”?


Intro
Deﬁne It
CF
Convergents
General

Deﬁne It

concept of a “convergent sum”? For a standard continued fraction
corresponding to {an }, let the partial convergent “stopping” at an
be the fraction hn/kn .


Intro
Deﬁne It
CF
Convergents
General

Deﬁne It

be the fraction hn/kn . As we’ll show, the series {hn } and {kn } are
very related and can be used to determine convergence.


Intro
Deﬁne It
CF
Convergents
General

Deﬁne It

Notice the zeroth partial is a0 so h0 = a0 , k0 = 1.


Intro
Define It
CF
Convergents
General

Define It

1 a1 (a0 )+1
The first is a0 + a1 = a1 (1)+0 , h1 = a1 a0 + 1, k1 = a1 .


Intro
Define It
CF
Convergents
General

Define It

1 a1 (a0 )+1
The first is a0 + a1 = a1 (1)+0 , h1 = a1 a0 + 1, k1 = a1 .
1 a2 a2 (a0 a1 +1)+a0
The second is a0 + = a0 + a1 a2 +1 = a2 (a1 )+1 .
1
a1 +
a2


Intro
Deﬁne It
CF
Convergents
General


Claim: hn and kn satisfy the same recurrence relation

hn = an hn−1 + hn−2
kn = an kn−1 + kn−2

along with initial conditions h−1 = 1, h−2 = 0 and
k−1 = 0, k−2 = 1.


Intro
Deﬁne It
CF
Convergents
General


Proof: There is one key insight: when go from one partial to the
1
1
next by turning the ﬁnal an into 1
. Using the recurrence
an + an+1
and the inductive assumption:


Intro
Deﬁne It
CF
Convergents
General


1
1
an + an+1
hn an hn−1 + hn−2
=
kn an kn−1 + kn−2


Intro
Deﬁne It
CF
Convergents
General


1
1
an + an+1
=
1
hn+1 an + an+1 hn−1 + hn−2
becomes =
kn+1 an + 1
kn−1 + kn−2
an+1


Intro
Deﬁne It
CF
Convergents
General


1
1
an + an+1
=
1
hn+1 an + an+1 hn−1 + hn−2
becomes =
kn+1 an + 1
kn−1 + kn−2
an+1

Why can we replace an straight-up?


Intro
Deﬁne It
CF
Convergents
General


1
1
an + an+1
=
1
hn+1 an + an+1 hn−1 + hn−2
becomes =
kn+1 an + 1
kn−1 + kn−2
an+1

Why can we replace an straight-up? All the other terms depend
solely on a1 , . . . , an−1 .

Intro
Deﬁne It
CF
Convergents
General

Standard

1
hn+1 an + an+1 hn−1 + hn−2
and =
kn+1 an + 1
kn−1 + kn−2
an+1


Intro
Deﬁne It
CF
Convergents
General

Standard

1
hn+1 an + an+1 hn−1 + hn−2
and =
kn+1 an + 1
kn−1 + kn−2
an+1
an+1 (an hn−1 + hn−2 ) + hn−1
=
an+1 (an kn−1 + kn−2 ) + kn−1


Intro
Deﬁne It
CF
Convergents
General

Standard

1
hn+1 an + an+1 hn−1 + hn−2
and =
kn+1 an + 1
kn−1 + kn−2
an+1
an+1 (an hn−1 + hn−2 ) + hn−1
=
an+1 (an kn−1 + kn−2 ) + kn−1
an+1 hn + hn−1
= by the inductive assumption.
an+1 kn + kn−1


Intro
Deﬁne It
CF
Convergents
General

Standard

1
hn+1 an + an+1 hn−1 + hn−2
and =
kn+1 an + 1
kn−1 + kn−2
an+1
an+1 (an hn−1 + hn−2 ) + hn−1
=
an+1 (an kn−1 + kn−2 ) + kn−1
an+1 hn + hn−1
= by the inductive assumption.
an+1 kn + kn−1

Clearly, the recurrence is determined by just two terms, so with a
simple check, h−1 = 1, h−2 = 0 and k−1 = 0, k−1 = 1
⇒ h0 = a0 , h1 = a0 a1 + 1 and k0 = 1, k1 = a1 as we’d wish.


Intro
Deﬁne It
CF
Convergents
General

General

If we look to our favorite Sloane number sequence (A001203), we
ﬁnd that standard continued fractions don’t do anything helpful
with π:


Intro
Deﬁne It
CF
Convergents
General

General

If we look to our favorite Sloane number sequence (A001203), we
ﬁnd that standard continued fractions don’t do anything helpful
with π:
1
π =3+
1
7+
1
15 +
1
1+
.
292 + . .


Intro
Deﬁne It
CF
Convergents
General

General

But...


Intro
Deﬁne It
CF
Convergents
General

General

But...
1
π =3+
9
6+
25
6+
49
6+
.
6 + ..


Intro
Deﬁne It
CF
Convergents
General

General

Let’s generalize our deﬁnition!


Intro
Deﬁne It
CF
Convergents
General

General
Too General?

Deﬁne a generalized continued fraction with two series of
nonnegative integers {an }∞ and {bn }∞ :
n=0 n=0


Intro
Deﬁne It
CF
Convergents
General

General
Too General?

Deﬁne a generalized continued fraction with two series of
nonnegative integers {an }∞ and {bn }∞ :
n=0 n=0

b1
a0 +
b2
a1 +
b3
a2 +
.
a3 + . .


Intro
Deﬁne It
CF
Convergents
General

General
Too General?

Again, let the partial convergent “stopping” at an be the fraction
hn/kn .


Intro
Deﬁne It
CF
Convergents
General

General
Too General?

hn/kn .


hn = an hn−1 + bn hn−2
kn = an kn−1 + bn kn−2

k−1 = 0, k−2 = 1.


Intro
Deﬁne It
CF
Convergents
General

General
Too General?

hn/kn .


hn = an hn−1 + bn hn−2
kn = an kn−1 + bn kn−2

k−1 = 0, k−2 = 1.

We won’t prove it, but the key insight is (you guessed it) turning
an into an + bn+1 .
a
n+1


Intro
CF Difference
Convergents Telescoping Partials

Introduction
The Wrong Way
Smart Men
Conway and Pell
Continued Fractions
Define It
General
Difference
Review
Smith Sum


Intro
CF Diﬀerence

Diﬀerence: Dn

In the actual talk, I displayed “Hey Dan, go back to the chalkboard
please.” and just talked to them. I have included it here since
there is no chalkboard.


Intro
CF Difference

Difference: Dn
hn
We wish to find the value of the continued fraction: lim .
n→∞ kn


Intro
CF Diﬀerence

Diﬀerence: Dn
hn
n→∞ kn
n−1
h1 hn hr hr +1
Notice − = −
k1 kn kr kr +1
r =1


Intro
CF Difference

Difference: Dn
hn
n→∞ kn
n−1
h1 hn hr hr +1
Notice − = − , with this it becomes clear it
k1 kn kr kr +1
r =1
is important to define the difference: Dn+1 = kn+1 hn − kn hn+1 .


Intro
CF Diﬀerence

Diﬀerence: Dn
hn
n→∞ kn
n−1
h1 hn hr hr +1
k1 kn kr kr +1
r =1

We have D1 = k1 h0 − k0 h1 = a1 a0 − 1 · (a1 a0 + b1 ) = −b1 and


Intro
CF Diﬀerence

Diﬀerence: Dn
hn
n→∞ kn
n−1
h1 hn hr hr +1
k1 kn kr kr +1
r =1

Dn+1 = (an+1 kn + bn+1 kn−1 )hn − kn (an+1 hn + bn+1 hn−1 )


Intro
CF Diﬀerence

Diﬀerence: Dn
hn
n→∞ kn
n−1
h1 hn hr hr +1
k1 kn kr kr +1
r =1

= bn+1 (kn−1 hn − kn hn−1 )


Intro
CF Diﬀerence

Diﬀerence: Dn
hn
n→∞ kn
n−1
h1 hn hr hr +1
k1 kn kr kr +1
r =1

= bn+1 (kn−1 hn − kn hn−1 )
= −bn+1 Dn


Intro
CF Diﬀerence

Diﬀerence: Dn
hn
n→∞ kn
n−1
h1 hn hr hr +1
k1 kn kr kr +1
r =1

= bn+1 (kn−1 hn − kn hn−1 )
= −bn+1 Dn
n
so any easy induction gives us Dn = (−1)n r =1 br .


Intro
CF Diﬀerence

Summary: Dn

After I returned from the chalkboard, I used this slide.


Intro
CF Diﬀerence

Summary: Dn

To summarize what we just said:

Dn+1 = kn+1 hn − kn hn+1
n
Dn = (−1)n br
r =1


Intro
CF Diﬀerence


Again, in the actual talk, I displayed “Hey Dan, go back to the
chalkboard please.” and just talked to them. I have included it
here since there is no chalkboard.


Intro
CF Diﬀerence

n
hn Dr
For the partials: = a0 −
kn kr kr −1
r =1


Intro
CF Diﬀerence

n
hn Dr
kn kr kr −1
r =1

n−1
h1 hn hr hr +1
− = −
k1 kn kr kr +1
r =1


Intro
CF Diﬀerence

n
hn Dr
kn kr kr −1
r =1

n−1
h1 hn hr hr +1
− = −
k1 kn kr kr +1
r =1
n−1
Dr +1
= but D1 = −b1 = a1 a0 − h1
kr kr +1
r =1


Intro
CF Diﬀerence

n
hn Dr
kn kr kr −1
r =1

n−1
h1 hn hr hr +1
− = −
k1 kn kr kr +1
r =1
n−1
Dr +1
= but D1 = −b1 = a1 a0 − h1
kr kr +1
r =1
n−1 n
hn h1 Dr +1 Dr
⇒ = − = a0 −
kn k1 kr kr +1 kr kr −1
r =1 r =1


Intro
CF Diﬀerence

n
hn Dr
kn kr kr −1
r =1

n−1
h1 hn hr hr +1
− = −
k1 kn kr kr +1
r =1
n−1
Dr +1
= but D1 = −b1 = a1 a0 − h1
kr kr +1
r =1
n−1 n
hn h1 Dr +1 Dr
⇒ = − = a0 −
kn k1 kr kr +1 kr kr −1
r =1 r =1

h1 a1 a0 −D1 a1 a0 D1
since k0 = 1, k1 = a1 ⇒ k1 = k1 = a1 − k1 k0 .


Intro
CF Diﬀerence

Summary: Partials



Intro
CF Diﬀerence

Summary: Partials

n
hn Dr
= a0 −
kn kr kr −1
r =1


Intro
CF Review
Convergents Smith Sum

Introduction
The Wrong Way
Smart Men
Conway and Pell
Continued Fractions
Deﬁne It
General
Diﬀerence
Review
Smith Sum


Intro
CF Review

Review

Recall that we showed:
∞
π 1 1 1 (−1)n
= 1 − + − + ··· =
4 3 5 7 2n + 1
n=0


Intro
CF Review

Smith Sum:
This is equivalent to the previous sum

Claim:
∞
π−3 (−1)n−1
=
4 (2n)(2n + 1)(2n + 2)
n=1


Intro
CF Review

Smith Sum:
This is equivalent to the previous sum

Claim:
∞
π−3 (−1)n−1
=
4 (2n)(2n + 1)(2n + 2)
n=1

Proof: Here we just manipulate the summand using the partial
fractal decomposition
1 1 1 1
= − + .
2n(2n + 1)(2n + 2) 4n 2n + 1 4n + 4


Intro
CF Review

Smith Sum:

∞
(−1)n−1
2n(2n + 1)(2n + 2)
n=1


Intro
CF Review

Smith Sum:

∞
(−1)n−1
2n(2n + 1)(2n + 2)
n=1
∞ ∞ ∞
(−1)n−1 (−1)n (−1)n−1
= + +
4n 2n + 1 4n + 4
n=1 n=1 n=1


Intro
CF Review

Smith Sum:

∞
(−1)n−1
2n(2n + 1)(2n + 2)
n=1
∞ ∞ ∞
(−1)n−1 (−1)n (−1)n−1
= + +
4n 2n + 1 4n + 4
n=1 n=1 n=1
∞ ∞ ∞
(−1)n−1 (−1)n (−1)n
= + −1+
4n 2n + 1 4n
n=1 n=0 n=2


Intro
CF Review

Smith Sum:

∞
(−1)n−1
2n(2n + 1)(2n + 2)
n=1
∞ ∞ ∞
(−1)n−1 (−1)n (−1)n−1
= + +
4n 2n + 1 4n + 4
n=1 n=1 n=1
∞ ∞ ∞
(−1)n−1 (−1)n (−1)n
= + −1+
4n 2n + 1 4n
n=1 n=0 n=2
(−1)1+1 π π−3
= + −1=
4·1 4 4


Parting Words

Part II

What We Came Here For: π


A Nifty Identity
Convergence
Parting Words
Plug and Play
Plug and Play

A Nifty Identity
Convergence
Plug and Play
Plug and Play

Parting Words
4 12
π and π 2
Thank You


A Nifty Identity
Convergence
Parting Words
Plug and Play
Plug and Play

A Nifty Identity

1 1 1 ∞ (−1)n−1
If c1 − c2 + c3 − ··· = n=1 cn converges then so does the
1
fraction 2
and they are
c1
c1 + 2
c2
c2 − c1 + 2
c3
c3 − c2 +
.
c4 − c3 + . .
equivalent, i.e. they are equal at every convergent.


A Nifty Identity
Convergence
Parting Words
Plug and Play
Plug and Play

A Nifty Identity

1 1 1 ∞ (−1)n−1
If c1 − c2 + c3 − ··· = n=1 cn converges then so does the
1
fraction 2
and they are
c1
c1 + 2
c2
c2 − c1 + 2
c3
c3 − c2 +
.
c4 − c3 + . .
equivalent, i.e. they are equal at every convergent.
Here a0 = 0, a1 = c1 , an = cn − cn−1 for n ≥ 2, b1 = 1, and
2
bn = cn−1 for n ≥ 2.


A Nifty Identity
Convergence
Parting Words
Plug and Play
Plug and Play

Denominator Series
n
Claim: kn = i=1 ci .


A Nifty Identity
Convergence
Parting Words
Plug and Play
Plug and Play

Denominator Series
Claim: kn = n ci .
i=1
Proof: We proceed by a strong induction. For n = 1, k1 = a1 = c1 ,
assuming it holds up to n = m for m ≥ 1


A Nifty Identity
Convergence
Parting Words
Plug and Play
Plug and Play

Denominator Series
Claim: kn = n ci .
i=1

km+1 = am+1 km + bm+1 km−1


A Nifty Identity
Convergence
Parting Words
Plug and Play
Plug and Play

Denominator Series
Claim: kn = n ci .
i=1

km+1 = am+1 km + bm+1 km−1
m m−1
2
= (cm+1 − cm ) ci + cm ci
i=1 i=1


A Nifty Identity
Convergence
Parting Words
Plug and Play
Plug and Play

Denominator Series
Claim: kn = n ci .
i=1

km+1 = am+1 km + bm+1 km−1
m m−1
2
i=1 i=1
m m
i=1 i=1


A Nifty Identity
Convergence
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Denominator Series
Claim: kn = n ci .
i=1

km+1 = am+1 km + bm+1 km−1
m m−1
2
i=1 i=1
m m
i=1 i=1
m+1
= ci . So the induction holds.
i=1


A Nifty Identity
Convergence
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Convergence
Claim:
∞
hn (−1)n−1
lim =
n→∞ kn cn
n=1


A Nifty Identity
Convergence
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Convergence
Claim:
∞
hn (−1)n−1
lim =
n→∞ kn cn
n=1
n n−1 2
Proof: We have Dn = (−1)n i=1 bi = (−1)n i=1 ci . Given
what we just proved about kn :


A Nifty Identity
Convergence
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Convergence
Claim:
∞
hn (−1)n−1
lim =
n→∞ kn cn
n=1
n n−1 2
n−1 2
Dn (−1)n i=1 ci (−1)n
= n n−1
=
kn kn−1 i=1 ci · i=1 ci
cn


A Nifty Identity
Convergence
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Convergence
Claim:
∞
hn (−1)n−1
lim =
n→∞ kn cn
n=1
n n−1 2
n−1 2
Dn (−1)n i=1 ci (−1)n
= n n−1
=
cn
∞
hn Dn
⇒ lim =0− since a0 = 0
n→∞ kn kn kn−1
n=1


A Nifty Identity
Convergence
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Convergence
Claim:
∞
hn (−1)n−1
lim =
n→∞ kn cn
n=1
n n−1 2
n−1 2
Dn (−1)n i=1 ci (−1)n
= n n−1
=
cn
∞
hn Dn
⇒ lim =0− since a0 = 0
n→∞ kn kn kn−1
n=1
∞ ∞
(−1)n (−1)n−1
=− =
cn cn
n=1 n=1


A Nifty Identity
Convergence
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One ﬁnal time in the actual talk, I displayed “Hey Dan, go back to
the chalkboard please.” and just talked to them. I have included it
here since there is no chalkboard.


A Nifty Identity
Convergence
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We can use this identity in a surprising way to ﬁnd the result of
this talk. The paper “An Elegant Continued Fraction for π” by
L.J. Lange credits one (presumably Douglas) D. Bowman for this
approach. Setting cn = 2n(2n + 1)(2n + 2) we have
a1 = c1 = 2 · 3 · 4 = 6(2 · 1)2 and for n ≥ 2,
an = 2n(2n + 1)(2n + 2) − (2n − 2)(2n − 1)2n = 24n2 = 6(2n)2 .


A Nifty Identity
Convergence
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bn
In any general continued fraction, the step can be
bn+1
an +
..
.
cbn
transformed to for any c ∈ R× without changing the
cbn+1
can +
..
.
value of the expression.


A Nifty Identity
Convergence
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bn
For n + 1 ≥ 2 we see the fraction =
bn+1
an +
..
.
(2n − 2)2 (2n − 1)2 (2n)2 (2n − 2)2 (2n − 1)2
= .
2+
(2n)2 (2n + 1)2 (2n + 2)2 (2n + 1)2 (2n + 2)2
6 · (2n) 6+
.. ..
. .
bn bn+1
So since bn+1 occurs in and , we cancel
bn+1 bn+2
an + an+1 +
.. ..
. .
(2n) 2 in the ﬁrst and (2n + 2)2 in the second. After this reduction,

each an is simply 6 and each bn+1 becomes (2n + 1)2 for n ≥ 2.


A Nifty Identity
Convergence
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Since b1 = 1 we are left to deal only with b2 . Our fraction has
become:
1 1
= .
22 32 42 4 · 32
6 · 22 + 6·4+
42 52 62 52
6 · 42 + 6+
. .
6 · 62 + . . 6 + ..


A Nifty Identity
Convergence
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Since b1 = 1 we are left to deal only with b2 . Our fraction has
become:
1 1
= .
22 32 42 4 · 32
6 · 22 + 6·4+
42 52 62 52
6 · 42 + 6+
. .
6 · 62 + . . 6 + ..
And from our identity, this fraction is equal to
∞ ∞
(−1)n−1 (−1)n−1 π−3
= = .
cn 2n(2n + 1)(2n + 2) 4
n=1 n=1


A Nifty Identity
Convergence
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This is the point where I returned from the chalkboard back to the
slides.


A Nifty Identity
Convergence
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π−3 1
=
4 4 · 32
6·4+
52
6+
.
6 + ..


A Nifty Identity
Convergence
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π−3 1 1 1
= =
4 4· 32 4 32
6·4+ 6+
52 52
6+ 6+
. .
6 + .. 6 + ..


A Nifty Identity
Convergence
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π−3 1 1 1
= =
4 4· 32 4 32
6·4+ 6+
52 52
6+ 6+
. .
6 + .. 6 + ..
1
⇐⇒ π = 3 +
32
6+
52
6+
.
6 + ..


Analysis and Arithmetic 4 and 12
π π2
Parting Words Thank You

A Nifty Identity
Convergence
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Parting Words
4 12
π and π 2
Thank You


π π2

4 12
π and π2

4 1
=1+
π 9
2+
25
2+
49
2+
.
2 + ..


π π2

4 12
π and π2

4 1
=1+
π 4
3+
9
5+
16
7+
.
9 + ..


π π2

4 12
π and π2

12 1
=1+
π2 16
3+
81
5+
256
7+
.
9 + ..


π π2

Thank You

Thank You for having me!


Life of π: Continued Fractions and Infinite Series

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