Life of π: Continued Fractions and Infinite Series

Outlines

Daniel J. Hermes

February 29, 2012

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

Outlines

π = 3 +1

6 +.. .

OutlinesPart I: Introductory FactsPart II: What We Came Here For: π

Outline of Part I

IntroductionThe Wrong WaySmart MenConway and Pell

Continued FractionsDefine ItWhat Does it Mean to Converge?General

Working with the ConvergentsDifferenceTelescoping Partials

Lemmata: Sums for πReviewSmith Sum

Outline of Part I

Outline of Part II

Analysis and ArithmeticA Nifty IdentityDenominator Series (kn)ConvergencePlug and PlayPlug and Play

Parting Words4π and 12

Thank You

Outline of Part II

Thank You

IntroCF

ConvergentsLemmata: Sums for π

Part I

Introductory Facts

IntroCF

The Wrong WaySmart MenConway and Pell

IntroCF

The Wrong Way

In bill #246 of the 1897 sitting of the Indiana General Assembly, πwas rational:

...Furthermore, it has revealed the ratio of the chord andarc of ninety degrees, which is as seven to eight, and alsothe ratio of the diagonal and one side of a square whichis as ten to seven, disclosing the fourth important fact,that the ratio of the diameter and circumference is asfive-fourths to four...

5/4= 3.2

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

IntroCF

The Wrong Way

5/4= 3.2

IntroCF

The Wrong Way

5/4= 3.2

IntroCF

Smart Men: Archimedes

In the third century BCE, Archimedes proved the sharpinequalities

223/71 < π < 22/7

by means of regular 96-gons

7= 3 +

IntroCF

Smart Men: Archimedes

In the third century BCE, Archimedes proved the sharpinequalities

223/71 < π < 22/7

by means of regular 96-gons

7= 3 +

IntroCF

Smart Men: The Bible

In 1 Kings 7:23 the word translated ’measuring line’appears in the Hebrew text ... The ratio of the numericalvalues of these Hebrew spellings is 111/106. If the putativevalue of 3 is multiplied by this ratio, one obtains 333/106

106= 3 +

= 3 +1

106/15

= 3 +1

IntroCF

106= 3 +

= 3 +1

106/15

= 3 +1

IntroCF

106= 3 +

= 3 +1

106/15

= 3 +1

IntroCF

106= 3 +

= 3 +1

106/15

= 3 +1

IntroCF

Smart Men: Zu Chongzhi

The 5th century Chinese mathematician and astronomerZu Chongzhi ... gave two other approximations ... 22/7and 355/113

113= 3 +

= 3 +1

113/16= 3 +

7 + 1/16

= 3 +1

15 + 1/1

IntroCF

113= 3 +

= 3 +1

113/16= 3 +

7 + 1/16

= 3 +1

15 + 1/1

IntroCF

113= 3 +

= 3 +1

113/16= 3 +

7 + 1/16

= 3 +1

15 + 1/1

IntroCF

113= 3 +

= 3 +1

113/16= 3 +

7 + 1/16

= 3 +1

15 + 1/1

IntroCF

Smart Men: James Gregory

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

arctan x = x − x3

5− x7

7+ · · ·

IntroCF

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

arctan x = x − x3

5− x7

7+ · · ·

IntroCF

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

4= 1− 1

5− 1

7+ · · · =

∞∑n=0

(−1)n

2n + 1

Proof: Since ddx arctan x =

1 + x2= 1− x2 + x4 − x6 + · · · and

arctan 0 = 0, integrating, we find the Taylor series expansion forarctan is

arctan x = x − x3

5− x7

7+ · · ·

Henceπ

4= arctan(1) =

∞∑n=0

(−1)n

2n + 1. �

IntroCF

4= 1− 1

5− 1

7+ · · · =

∞∑n=0

(−1)n

2n + 1

1 + x2= 1− x2 + x4 − x6 + · · · and

arctan x = x − x3

5− x7

7+ · · ·

Henceπ

4= arctan(1) =

∞∑n=0

(−1)n

2n + 1. �

IntroCF

4= 1− 1

5− 1

7+ · · · =

∞∑n=0

(−1)n

2n + 1

1 + x2= 1− x2 + x4 − x6 + · · · and

arctan x = x − x3

5− x7

7+ · · ·

Henceπ

4= arctan(1) =

∞∑n=0

(−1)n

2n + 1. �

IntroCF

4= 1− 1

5− 1

7+ · · · =

∞∑n=0

(−1)n

2n + 1

1 + x2= 1− x2 + x4 − x6 + · · · and

arctan x = x − x3

5− x7

7+ · · ·

Henceπ

4= arctan(1) =

∞∑n=0

(−1)n

2n + 1. �

IntroCF

4= 1− 1

5− 1

7+ · · · =

∞∑n=0

(−1)n

2n + 1

1 + x2= 1− x2 + x4 − x6 + · · · and

arctan x = x − x3

5− x7

7+ · · ·

Henceπ

4= arctan(1) =

∞∑n=0

(−1)n

2n + 1. �

IntroCF

Conway and Pell

In the actual talk, I displayed “Hey Dan, go to the chalkboardplease.” and just talked to them. For some reference see:

I blog.bossylobster.com/2011/07/continued-fractions-for-greater-good.html

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

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

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

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

IntroCF

Define ItWhat Does it Mean to Converge?General

IntroCF

Define It

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

a3 +.. .

IntroCF

Define It

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

a3 +.. .

IntroCF

Define It

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

For a standard continued fractioncorresponding to {an}, let the partial convergent “stopping” at anbe the fraction hn/kn. As we’ll show, the series {hn} and {kn} arevery related and can be used to determine convergence.

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

I The first is a0 + 1a1

= a1(a0)+1a1(1)+0 , h1 = a1a0 + 1, k1 = a1.

I The second is a0 +1

= a0 + a2a1a2+1 = a2(a0a1+1)+a0

a2(a1)+1 .

IntroCF

Define It

How will we know if our fractions converge if we don’t have aconcept of a “convergent sum”? For a standard continued fractioncorresponding to {an}, let the partial convergent “stopping” at anbe the fraction hn/kn.

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

= a1(a0)+1a1(1)+0 , h1 = a1a0 + 1, k1 = a1.

= a0 + a2a1a2+1 = a2(a0a1+1)+a0

a2(a1)+1 .

IntroCF

Define It

How will we know if our fractions converge if we don’t have aconcept of a “convergent sum”? For a standard continued fractioncorresponding to {an}, let the partial convergent “stopping” at anbe the fraction hn/kn. As we’ll show, the series {hn} and {kn} arevery related and can be used to determine convergence.

= a1(a0)+1a1(1)+0 , h1 = a1a0 + 1, k1 = a1.

= a0 + a2a1a2+1 = a2(a0a1+1)+a0

a2(a1)+1 .

IntroCF

Define It

= a1(a0)+1a1(1)+0 , h1 = a1a0 + 1, k1 = a1.

= a0 + a2a1a2+1 = a2(a0a1+1)+a0

a2(a1)+1 .

IntroCF

Define It

= a1(a0)+1a1(1)+0 , h1 = a1a0 + 1, k1 = a1.

= a0 + a2a1a2+1 = a2(a0a1+1)+a0

a2(a1)+1 .

IntroCF

Define It

= a1(a0)+1a1(1)+0 , h1 = a1a0 + 1, k1 = a1.

= a0 + a2a1a2+1 = a2(a0a1+1)+a0

a2(a1)+1 .

IntroCF

What Does it Mean to Converge?

Claim: hn and kn satisfy the same recurrence relation

hn = anhn−1 + hn−2

kn = ankn−1 + kn−2

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

IntroCF

Proof: There is one key insight: when go from one partial to the

next by turning the final 1an

an + 1an+1

. Using the recurrence

and the inductive assumption:

anhn−1 + hn−2

ankn−1 + kn−2

becomeshn+1

(an + 1

)hn−1 + hn−2(

an + 1an+1

)kn−1 + kn−2

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

IntroCF

an + 1an+1

anhn−1 + hn−2

ankn−1 + kn−2

becomeshn+1

(an + 1

)hn−1 + hn−2(

an + 1an+1

)kn−1 + kn−2

IntroCF

an + 1an+1

anhn−1 + hn−2

ankn−1 + kn−2

becomeshn+1

(an + 1

)hn−1 + hn−2(

an + 1an+1

)kn−1 + kn−2

IntroCF

an + 1an+1

anhn−1 + hn−2

ankn−1 + kn−2

becomeshn+1

(an + 1

)hn−1 + hn−2(

an + 1an+1

)kn−1 + kn−2

Why can we replace an straight-up?

All the other terms dependsolely on a1, . . . , an−1.

IntroCF

an + 1an+1

anhn−1 + hn−2

ankn−1 + kn−2

becomeshn+1

(an + 1

)hn−1 + hn−2(

an + 1an+1

)kn−1 + kn−2

IntroCF

Standard

andhn+1

(an + 1

)hn−1 + hn−2(

an + 1an+1

)kn−1 + kn−2

=an+1(anhn−1 + hn−2) + hn−1

an+1(ankn−1 + kn−2) + kn−1

=an+1hn + hn−1

an+1kn + kn−1by the inductive assumption.

Clearly, the recurrence is determined by just two terms, so with asimple check, h−1 = 1, h−2 = 0 and k−1 = 0, k−1 = 1⇒ h0 = a0, h1 = a0a1 + 1 and k0 = 1, k1 = a1 as we’d wish. �

IntroCF

Standard

andhn+1

(an + 1

)hn−1 + hn−2(

an + 1an+1

)kn−1 + kn−2

=an+1(anhn−1 + hn−2) + hn−1

an+1(ankn−1 + kn−2) + kn−1

=an+1hn + hn−1

IntroCF

Standard

andhn+1

(an + 1

)hn−1 + hn−2(

an + 1an+1

)kn−1 + kn−2

=an+1(anhn−1 + hn−2) + hn−1

an+1(ankn−1 + kn−2) + kn−1

=an+1hn + hn−1

IntroCF

Standard

andhn+1

(an + 1

)hn−1 + hn−2(

an + 1an+1

)kn−1 + kn−2

=an+1(anhn−1 + hn−2) + hn−1

an+1(ankn−1 + kn−2) + kn−1

=an+1hn + hn−1

IntroCF

General

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

π = 3 +1

292 +. . .

IntroCF

General

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

π = 3 +1

292 +. . .

IntroCF

General

But...

π = 3 +1

6 +.. .

IntroCF

General

But...

π = 3 +1

6 +.. .

IntroCF

General

Let’s generalize our definition!

IntroCF

GeneralToo General?

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

a0 +b1

a1 +b2

a2 +b3

a3 +.. .

IntroCF

GeneralToo General?

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

a0 +b1

a1 +b2

a2 +b3

a3 +.. .

IntroCF

GeneralToo General?

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

hn = anhn−1 + bnhn−2

kn = ankn−1 + bnkn−2

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

IntroCF

GeneralToo General?

IntroCF

GeneralToo General?

IntroCF

DifferenceTelescoping Partials

IntroCF

Difference: Dn

In the actual talk, I displayed “Hey Dan, go back to the chalkboardplease.” and just talked to them. I have included it here since

there is no chalkboard.

IntroCF

Difference: Dn

We wish to find the value of the continued fraction: limn→∞

Noticeh1

k1− hn

n−1∑r=1

kr− hr+1

), with this it becomes clear it

is important to define the difference: Dn+1 = kn+1hn − knhn+1.

We have D1 = k1h0 − k0h1 = a1a0 − 1 · (a1a0 + b1) = −b1 and

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

= bn+1(kn−1hn − knhn−1)

= −bn+1Dn

so any easy induction gives us Dn = (−1)n∏n

r=1 br .

IntroCF

Difference: Dn

Noticeh1

k1− hn

n−1∑r=1

kr− hr+1

, with this it becomes clear it

= −bn+1Dn

r=1 br .

IntroCF

Difference: Dn

Noticeh1

k1− hn

n−1∑r=1

kr− hr+1

= −bn+1Dn

r=1 br .

IntroCF

Difference: Dn

Noticeh1

k1− hn

n−1∑r=1

kr− hr+1

= −bn+1Dn

r=1 br .

IntroCF

Difference: Dn

Noticeh1

k1− hn

n−1∑r=1

kr− hr+1

= −bn+1Dn

r=1 br .

IntroCF

Difference: Dn

Noticeh1

k1− hn

n−1∑r=1

kr− hr+1

= −bn+1Dn

r=1 br .

IntroCF

Difference: Dn

Noticeh1

k1− hn

n−1∑r=1

kr− hr+1

= −bn+1Dn

r=1 br .

IntroCF

Difference: Dn

Noticeh1

k1− hn

n−1∑r=1

kr− hr+1

= −bn+1Dn

r=1 br .

IntroCF

Summary: Dn

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

To summarize what we just said:

Dn+1 = kn+1hn − knhn+1

Dn = (−1)nn∏

IntroCF

Summary: Dn

After I returned from the chalkboard, I used this slide.To summarize what we just said:

Dn+1 = kn+1hn − knhn+1

Dn = (−1)nn∏

IntroCF

Telescoping Partials

Again, in the actual talk, I displayed “Hey Dan, go back to thechalkboard please.” and just talked to them. I have included it

here since there is no chalkboard.

IntroCF

For the partials:hn

kn= a0 −

n∑r=1

krkr−1

k1− hn

n−1∑r=1

kr− hr+1

=n−1∑r=1

krkr+1but D1 = −b1 = a1a0 − h1

⇒ hn

n−1∑r=1

krkr+1= a0 −

n∑r=1

krkr−1

since k0 = 1, k1 = a1 ⇒ h1k1

= a1a0−D1k1

= a1a0a1− D1

IntroCF

For the partials:hn

kn= a0 −

n∑r=1

krkr−1

k1− hn

n−1∑r=1

kr− hr+1

=n−1∑r=1

⇒ hn

n−1∑r=1

krkr+1= a0 −

n∑r=1

krkr−1

since k0 = 1, k1 = a1 ⇒ h1k1

= a1a0−D1k1

= a1a0a1− D1

IntroCF

For the partials:hn

kn= a0 −

n∑r=1

krkr−1

k1− hn

n−1∑r=1

kr− hr+1

=n−1∑r=1

⇒ hn

n−1∑r=1

krkr+1= a0 −

n∑r=1

krkr−1

since k0 = 1, k1 = a1 ⇒ h1k1

= a1a0−D1k1

= a1a0a1− D1

IntroCF

For the partials:hn

kn= a0 −

n∑r=1

krkr−1

k1− hn

n−1∑r=1

kr− hr+1

=n−1∑r=1

⇒ hn

n−1∑r=1

krkr+1= a0 −

n∑r=1

krkr−1

since k0 = 1, k1 = a1 ⇒ h1k1

= a1a0−D1k1

= a1a0a1− D1

IntroCF

For the partials:hn

kn= a0 −

n∑r=1

krkr−1

k1− hn

n−1∑r=1

kr− hr+1

=n−1∑r=1

⇒ hn

n−1∑r=1

krkr+1= a0 −

n∑r=1

krkr−1

since k0 = 1, k1 = a1 ⇒ h1k1

= a1a0−D1k1

= a1a0a1− D1

IntroCF

Summary: Partials

kn= a0 −

n∑r=1

krkr−1

IntroCF

Summary: Partials

kn= a0 −

n∑r=1

krkr−1

IntroCF

ReviewSmith Sum

IntroCF

ReviewSmith Sum

Review

Recall that we showed:

4= 1− 1

5− 1

7+ · · · =

∞∑n=0

(−1)n

2n + 1

IntroCF

ReviewSmith Sum

Smith Sum:This is equivalent to the previous sum

Claim:π − 3

4=∞∑n=1

(−1)n−1

(2n)(2n + 1)(2n + 2)

Proof: Here we just manipulate the summand using the partialfractal decomposition

2n(2n + 1)(2n + 2)=

4n− 1

2n + 1+

4n + 4.

IntroCF

ReviewSmith Sum

Smith Sum:This is equivalent to the previous sum

Claim:π − 3

4=∞∑n=1

(−1)n−1

(2n)(2n + 1)(2n + 2)

Proof: Here we just manipulate the summand using the partialfractal decomposition

2n(2n + 1)(2n + 2)=

4n− 1

2n + 1+

4n + 4.

IntroCF

ReviewSmith Sum

Smith Sum:

∞∑n=1

(−1)n−1

2n(2n + 1)(2n + 2)

=∞∑n=1

(−1)n−1

4n+∞∑n=1

(−1)n

2n + 1+∞∑n=1

(−1)n−1

4n + 4

=∞∑n=1

(−1)n−1

4n+∞∑n=0

(−1)n

2n + 1− 1 +

∞∑n=2

(−1)n

=(−1)1+1

4 · 1+π

4− 1 =

π − 3

IntroCF

ReviewSmith Sum

Smith Sum:

∞∑n=1

(−1)n−1

2n(2n + 1)(2n + 2)

=∞∑n=1

(−1)n−1

4n+∞∑n=1

(−1)n

2n + 1+∞∑n=1

(−1)n−1

4n + 4

=∞∑n=1

(−1)n−1

4n+∞∑n=0

(−1)n

2n + 1− 1 +

∞∑n=2

(−1)n

=(−1)1+1

4 · 1+π

4− 1 =

π − 3

IntroCF

ReviewSmith Sum

Smith Sum:

∞∑n=1

(−1)n−1

2n(2n + 1)(2n + 2)

=∞∑n=1

(−1)n−1

4n+∞∑n=1

(−1)n

2n + 1+∞∑n=1

(−1)n−1

4n + 4

=∞∑n=1

(−1)n−1

4n+∞∑n=0

(−1)n

2n + 1− 1 +

∞∑n=2

(−1)n

=(−1)1+1

4 · 1+π

4− 1 =

π − 3

IntroCF

ReviewSmith Sum

Smith Sum:

∞∑n=1

(−1)n−1

2n(2n + 1)(2n + 2)

=∞∑n=1

(−1)n−1

4n+∞∑n=1

(−1)n

2n + 1+∞∑n=1

(−1)n−1

4n + 4

=∞∑n=1

(−1)n−1

4n+∞∑n=0

(−1)n

2n + 1− 1 +

∞∑n=2

(−1)n

=(−1)1+1

4 · 1+π

4− 1 =

π − 3

Analysis and ArithmeticParting Words

Part II

What We Came Here For: π

A Nifty IdentityDenominator Series (kn)ConvergencePlug and PlayPlug and Play

Thank You

A Nifty Identity

If 1c1− 1

c3− · · · =

∑∞n=1

(−1)n−1

cnconverges then so does the

fraction1

c1 +c2

c2 − c1 +c2

c3 − c2 +c2

c4 − c3 +.. .

and they are

equivalent, i.e. they are equal at every convergent.

Here a0 = 0, a1 = c1, an = cn − cn−1 for n ≥ 2, b1 = 1, andbn = c2

n−1 for n ≥ 2.

A Nifty Identity

If 1c1− 1

c3− · · · =

∑∞n=1

(−1)n−1

cnconverges then so does the

fraction1

c1 +c2

c2 − c1 +c2

c3 − c2 +c2

c4 − c3 +.. .

and they are

equivalent, i.e. they are equal at every convergent.Here a0 = 0, a1 = c1, an = cn − cn−1 for n ≥ 2, b1 = 1, andbn = c2

n−1 for n ≥ 2.

Denominator Series

Claim: kn =∏n

i=1 ci .

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

km+1 = am+1km + bm+1km−1

= (cm+1 − cm)m∏i=1

ci + c2m

m−1∏i=1

= (cm+1 − cm)m∏i=1

ci + cm

m∏i=1

=m+1∏i=1

ci . So the induction holds. �

Denominator Series

Claim: kn =∏n

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

= (cm+1 − cm)m∏i=1

ci + c2m

m−1∏i=1

= (cm+1 − cm)m∏i=1

ci + cm

m∏i=1

=m+1∏i=1

Denominator Series

Claim: kn =∏n

= (cm+1 − cm)m∏i=1

ci + c2m

m−1∏i=1

= (cm+1 − cm)m∏i=1

ci + cm

m∏i=1

=m+1∏i=1

Denominator Series

Claim: kn =∏n

= (cm+1 − cm)m∏i=1

ci + c2m

m−1∏i=1

= (cm+1 − cm)m∏i=1

ci + cm

m∏i=1

=m+1∏i=1

Denominator Series

Claim: kn =∏n

= (cm+1 − cm)m∏i=1

ci + c2m

m−1∏i=1

= (cm+1 − cm)m∏i=1

ci + cm

m∏i=1

=m+1∏i=1

Denominator Series

Claim: kn =∏n

= (cm+1 − cm)m∏i=1

ci + c2m

m−1∏i=1

= (cm+1 − cm)m∏i=1

ci + cm

m∏i=1

=m+1∏i=1

Convergence

Claim:

limn→∞

kn=∞∑n=1

(−1)n−1

Proof: We have Dn = (−1)n∏n

i=1 bi = (−1)n∏n−1

i=1 c2i . Given

what we just proved about kn:

knkn−1=

(−1)n∏n−1

i=1 c2i∏n

i=1 ci ·∏n−1

i=1 ci=

(−1)n

⇒ limn→∞

kn= 0−

∞∑n=1

knkn−1since a0 = 0

= −∞∑n=1

(−1)n

cn=∞∑n=1

(−1)n−1

Convergence

Claim:

limn→∞

kn=∞∑n=1

(−1)n−1

i=1 bi = (−1)n∏n−1

i=1 c2i . Given

knkn−1=

(−1)n∏n−1

i=1 c2i∏n

i=1 ci ·∏n−1

i=1 ci=

(−1)n

⇒ limn→∞

kn= 0−

∞∑n=1

= −∞∑n=1

(−1)n

cn=∞∑n=1

(−1)n−1

Convergence

Claim:

limn→∞

kn=∞∑n=1

(−1)n−1

i=1 bi = (−1)n∏n−1

i=1 c2i . Given

knkn−1=

(−1)n∏n−1

i=1 c2i∏n

i=1 ci ·∏n−1

i=1 ci=

(−1)n

⇒ limn→∞

kn= 0−

∞∑n=1

= −∞∑n=1

(−1)n

cn=∞∑n=1

(−1)n−1

Convergence

Claim:

limn→∞

kn=∞∑n=1

(−1)n−1

i=1 bi = (−1)n∏n−1

i=1 c2i . Given

knkn−1=

(−1)n∏n−1

i=1 c2i∏n

i=1 ci ·∏n−1

i=1 ci=

(−1)n

⇒ limn→∞

kn= 0−

∞∑n=1

= −∞∑n=1

(−1)n

cn=∞∑n=1

(−1)n−1

Convergence

Claim:

limn→∞

kn=∞∑n=1

(−1)n−1

i=1 bi = (−1)n∏n−1

i=1 c2i . Given

knkn−1=

(−1)n∏n−1

i=1 c2i∏n

i=1 ci ·∏n−1

i=1 ci=

(−1)n

⇒ limn→∞

kn= 0−

∞∑n=1

= −∞∑n=1

(−1)n

cn=∞∑n=1

(−1)n−1

Plug and Play

One final time in the actual talk, I displayed “Hey Dan, go back tothe chalkboard please.” and just talked to them. I have included it

here since there is no chalkboard.

Plug and Play

We can use this identity in a surprising way to find the result ofthis talk. The paper “An Elegant Continued Fraction for π” byL.J. Lange credits one (presumably Douglas) D. Bowman for thisapproach. Setting cn = 2n(2n + 1)(2n + 2) we havea1 = 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.

Plug and Play

In any general continued fraction, the stepbn

an +bn+1

can be

transformed tocbn

can +cbn+1

for any c ∈ R× without changing the

value of the expression.

Plug and Play

For n + 1 ≥ 2 we see the fractionbn

an +bn+1

(2n − 2)2(2n − 1)2(2n)2

6 · (2n)2 +(2n)2(2n + 1)2(2n + 2)2

=(2n − 2)2(2n − 1)2

6 +(2n + 1)2(2n + 2)2

So since bn+1 occurs inbn

an +bn+1

andbn+1

an+1 +bn+2

, we cancel

(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.

Plug and Play

Since b1 = 1 we are left to deal only with b2. Our fraction hasbecome:

6 · 22 +223242

6 · 42 +425262

6 · 62 +.. .

6 · 4 +4 · 32

6 +. . .

And from our identity, this fraction is equal to

∞∑n=1

(−1)n−1

cn=∞∑n=1

(−1)n−1

2n(2n + 1)(2n + 2)=π − 3

Plug and Play

Since b1 = 1 we are left to deal only with b2. Our fraction hasbecome:

6 · 22 +223242

6 · 42 +425262

6 · 62 +.. .

6 · 4 +4 · 32

6 +. . .

And from our identity, this fraction is equal to

∞∑n=1

(−1)n−1

cn=∞∑n=1

(−1)n−1

2n(2n + 1)(2n + 2)=π − 3

Plug and Play

This is the point where I returned from the chalkboard back to theslides.

Plug and Play

π − 3

6 · 4 +4 · 32

6 +. . .

⇐⇒ π = 3 +1

6 +. . .

Plug and Play

π − 3

6 · 4 +4 · 32

6 +. . .

⇐⇒ π = 3 +1

6 +. . .

Plug and Play

π − 3

6 · 4 +4 · 32

6 +. . .

⇐⇒ π = 3 +1

6 +. . .

and 12π2

Thank You

and 12π2

Thank You

4π and 12

π= 1 +

2 +.. .

and 12π2

Thank You

4π and 12

π= 1 +

9 +.. .

and 12π2

Thank You

4π and 12

π2= 1 +

7 +256

9 +. . .

and 12π2

Thank You

Thank You for having me!

Life of π: Continued Fractions and Infinite Series

Education

Transcript of Life of π: Continued Fractions and Infinite Series

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