On expressing $\frac{\pi^n}{4\cdot 3^{n-1}}$ as a continued fraction.

Question

It is a celebrated equation that $$\frac{\pi}{4}=\cfrac{1}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\ddots}}}}$$

However, there are two other conjectured equations that I found which, if true (they seem to be), might reveal a pattern.

$$\frac{\pi^2}{12}=\cfrac{1}{1+\cfrac{1^4}{3+\cfrac{2^4}{5+\cfrac{3^4}{7+\ddots}}}}$$

$$\frac{\pi^3}{36}=\cfrac{1}{1+\cfrac{1^6}{3+\cfrac{2^6}{5+\cfrac{3^6}{7+\ddots}}}}$$

Conjectured General Formula: For natural $n\geqslant 1$, $$\frac{\pi^n}{4\cdot 3^{n-1}}=\cfrac{1}{1+\cfrac{1^{2n}}{3+\cfrac{2^{2n}}{5+\cfrac{3^{2n}}{7+\ddots}}}}$$

Can these be numerically verified? I have not the skill to by-hand prove/disprove these, and have only been using Wolfram Alpha to arrive at these conjectures.

It would also be much appreciated if one could suggest a program I could install in order to evaluate these continued fractions independently, as well as the code required. Will PARI/GP suffice?

Thanks.

Answer 1

There is a continued fraction in "Ramanujan’s Continued Fractions, Apéry’s Constant, and More" by Tito Piezas III from "A Collection of Algebraic Identities":

Entry 30:

$$ f(x) = \sum_{k=0}^\infty\frac1{(x+2k+1)^2} =\cfrac{1}{x+} \;\cfrac{1^4}{3x+} \;\cfrac{2^4}{5x+} \;\cfrac{3^4}{7x+} \;\cdots $$

which yields $f(1)=\pi^2/12$. There is also a route in "Entry 16."

Please notice that I didn't read that paper and I am just citing, and I don't know if it is proven or if it just follows from a conjecture.

However, the alleged fraction for π³/36 diverges, and here is why. $\def\K#1#2#3#4{\underset{#1}{\overset{#2}{\operatorname K}} \frac{#3}{#4}}$ Let's consider continued fractions of the form $$ f(k)=\K{n=1}{\infty}{a_n^k}{b_n} $$ with $a_n=n$ and $b_n=2n+1$. Your assertion is then expressed as $$ \frac{\pi^3}{36} \stackrel ?= \cfrac{1}{1+f(6)} $$ where the right side converges iff $f(6)$ converges. Now divide all partial fractions by their numerator which gets a new representation with the same convergence behavior: $$ f(k)=\K{n=1}{\infty}{1}{c_n^k b_n} $$ where the $c_n$ satisfy the recurrence $c_n=1/(a_n c_{n-1})$. With the above definition of $a_n$, this gives the explicit representation $$ c_n =\frac{(n-1)!!}{n!!} \approx \sqrt\frac{2}{\pi n} $$ where $!!$ denotes the double factorial. The approximation follows from properties of the Γ function and can be expressed less sloppily, in particular $$ \lim_{n\to\infty}\frac{\Gamma (n + \alpha)}{n^\alpha\Gamma(n)} = 1 \qquad\text{ applied to }\qquad n!! = \sqrt{\frac{2^{n+1}}{\pi}} \Gamma\left(\tfrac{n}{2}+1\right) $$ Then observe that the series $$ \sum_n c_n^k b_n \approx \left(\frac 2\pi\right)^{k/2} \sum_n n^{-k/2}(2n+1) $$ with $k$ fixed converges absolutely for $1-k/2 < -1 \Leftrightarrow k > 4$. From the absolute convergence of that series

it follows that $f(k)$ diverges by oscillation for $k>4$.

Addendum: For your future research we get the following take-away: Denote $$ f_n \sim g_n \Longleftrightarrow \left( g_n\to\infty \text{ and } 0 < \liminf_{n\to\infty}\,\frac{f_n}{g_n} \leqslant \limsup_{n\to\infty}\,\frac{f_n}{g_n} < \infty \right) $$ Using that notation, we get the corollary

Let $a_n, b_n>0$ be two sequences with $a_n\sim n^\alpha$ and $b_n\sim n^\beta$. Then $$ \alpha - 2\beta > 2 \quad\Longrightarrow\quad \K{n=1}{\infty}{a_n}{b_n} \;\text{ diverges by oscillation.} $$

(Notice that the notation for ~ implies α, β > 0.)

Answer 2

I cannot confirm the last fraction for π³/36. (Addendum: See my other answer for why that fraction does diverge (by oscillation b.t.w.)).

I used the following quick Python script. math.pi is only of limited precision, but it's error is much smaller than the error for the 3rd fraction.

#!/usr/bin/env python

from __future__ import print_function
from decimal import *
from math import *

getcontext().prec = 1000

def expand (N, expo):
    f = Decimal(0)
    for i in range (N, 0, -1):
        z = Decimal(i) ** expo
        n = 2*i + 1
        print ("%d / %d+" % (z, n))
        f = z / (n + f)
    print ("%d / %d+" % (1, 1))
    return 1 / (1 + f)

Pi = Decimal(pi)
res = expand (1200, 6)
print ("frac is approx", res)
print ("\nfrac - pi^3 / 36  is approx: ", res - Pi*Pi*Pi / 36)
print ("\n'float' precision is roughly 2^{%f}"
       % (log(abs(sqrt(pi) ** 2 - pi)) / log(2)))

Running:

...
15625 / 11+
4096 / 9+
729 / 7+
64 / 5+
1 / 3+
1 / 1+
frac is approx 0.9080445148...
frac - pi^3 / 36  is approx:  0.0467590515...
'float' precision is roughly 2^{-51.000000}

The last is what is expected from IEEE double precision, i.e. the error in Pi^3 / 36 is of similar magnitude and orders of magnitude smaller than 0.046.

Note: This is just an indicator in which direction to go; I didn't bother with how fast / good that fraction converges and how errors propagate. (One could use Fraction (rationals) but they will explode.)

Answer 3

Using Pari/GP with 200 dec digits internal precision and up to 2 million entries for the continued fractions the results are inconclusive.

• Preface: I'm reformulating your continued fractions slightly to match my Pari/GP-procedures. Instead of $$\Tiny \frac{\pi}{4}=\cfrac{1}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\ddots}}}}$$ I use $$ \Tiny w_0={1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\ddots}}}} $$ $$ w_1=1/w_0 \\ \frac{\pi}{4}=w_1$$

For $\pi/4$ and $\pi^2/12$ the results approximate well, the latter needs many coefficients for the continued fraction.

For $\pi^3/36$ the results are inconclusive. The approximations for $k,k+2,k+4,...$ coefficients seem to converge to one value, and the approximations for $k+1,k+3,k+5,...$ coefficients seem to converge to another value, so that possibly the continued fraction as such might not be convergent.

Below the protocols for the $\pi/4$ , $\pi^2/12$ and the $\pi^3/36$ approaches.

I have the definition for the entries of the cont-fraction as strings in a function call, plus the number of coefficients to be used.

• Setting for $\pi/4$
  w0 = CF("1+2*k", "(1+k)^2", 2000 ); w1=1/w0; print([w1,w2=Pi/4,err=w1-w2])
  gives [0.785398163397, 0.785398163397, 5.613524196 E-202]
  correct result to 202 dec digits.

• Setting for $\pi^2/12$
  w0 = CF("1+2*k", "(1+k)^4", 20 000 ); w1=1/w0; print([w1,w2=Pi^2/12,err=w1-w2])
w0 = CF("1+2*k", "(1+k)^4", 20 001 ); w1=1/w0; print([w1,w2=Pi^2/12,err=w1-w2])
  gives
  [0.82246703 4674, 0.822467033424, 0.00000000124981251875]
[0.82246703 2174, 0.822467033424, -0.00000000124968755624]
  correct result to 10 dec digits. But note, that the continued fraction is not yet in a usable state of approximation!
  Let's use 200 000 and 200 001 coefficients:
  [0.8224670334 37, 0.822467033424, 1.24998125019 E-11]
[0.8224670334 12, 0.822467033424, -1.24996875056 E-11]
  The convergence needs exorbitant many coefficients to achieve usable approximation, but at least it seems to work in general.

• Setting for $\pi^3/36$
  w0 = CF("1+2*k", "(1+k)^6", 2 000 000 ); w1=1/w0; print([w1,w2=Pi^3/36,err=w1-w2])
w0 = CF("1+2*k", "(1+k)^6", 2 000 001 ); w1=1/w0; print([w1,w2=Pi^3/36,err=w1-w2])
  gives
  [0.907969268604, 0.861285463342, 0.0466838052623]
[0.810934471732, 0.861285463342, -0.0503509916094]
  and giving more coefficients becomes usesless in practical purposes.

Conclusion: the first two continued fractions seem to be correct, because of the improvement of accuracy roughly related to the increase of number-of-used coefficients.
With the third continued fraction I've doubts, whether it might be possible that the continued fraction cannot converge at all. But I've never read such a thing - on the other hand I don't know much about non-simple continued fractions like in your cases here. Added: the wikipedia-entry on generalized continued fractions mentions the possibility of non-convergence, so this may well happen here.
But if it can be proved that it converges then I'd guess that it converges to your expected value.
Generalization: I think in general numerical checks on the common road of computation become useless and it is needed to apply algebraic arguments and find an analytical answer.

appendix: Pari/GP-program used.

  {CF(funcA,funcB="1",maxcoeffs=100) = local(convgts,a_k,b_k);
    eval(Str("A_(k)=",funcA));  \\ installing the function A_(n) given by string funcA
    eval(Str("B_(k)=",funcB));  \\ installing the function B_(n) given by string funcB
    convgts = [B_(0),A_(0);0,1];  \\ initialize
      for(k=1,maxcoeffs, 
                a_k = A_(k);
                b_k = B_(k); 
                convgts = convgts * [0,1; b_k,a_k]; \\matrix-multiplication
         );
    return( convgts[1,2]/ convgts[2,2]*1.0) ; }

Example: CF("1+k","1") evaluates the generalized continued fraction using A=[1;2,3,4,5,...] , B=[1,1,1,1,1,...] and because B has only ones as coefficients this is a simple continued fraction.

Example: CF("1+k","2+k") evaluates the generalized continued fraction using A=[1;2,3,4,5,...] , B=[2,3,4,5,6,...] giving $\exp(1)-1 \approx 1.71828$ .

Note, for the reference to the index of the coefficient use the symbol k in your formula-string. The first entry is indexed by k=0.

update
Here a protocol for the approximation-history of the $\pi^3/36$-problem. The odd and even indexed convergents from 20001 on are displayed:

   odd convergents           even convergents
   k     cvgt(k)          k     cvgt(k) 
  --------------------------------------------
  20001  1.23315310020  20002  1.10135343484
  20003  1.23315309941  20004  1.10135343539
  20005  1.23315309862  20006  1.10135343594
  20007  1.23315309783  20008  1.10135343648
  20009  1.23315309704  20010  1.10135343703
  20011  1.23315309625  20012  1.10135343758
  20013  1.23315309545  20014  1.10135343813
  20015  1.23315309466  20016  1.10135343867
  20017  1.23315309387  20018  1.10135343922
  20019  1.23315309308  20020  1.10135343977

We see the micro-minimal improve towards a common convergent downwards in the left table and upwards in the right table.
Here the binomial-transforms (P^-1*List)

        inverse binomial transform of the list
   odd convergents                     even convergents
   k     cvgt(k)                     k     cvgt(k) 
  ---------------------------------------------------------------
  20001             1.23315310020  20002            1.10135343484
      2  -0.000000000791808010433      2  0.000000000548026207045
      0         1.58319661722E-13      0       -1.09568710520E-13
      0        -4.74785730458E-17      0        3.28563729351E-17
      0         1.89826039525E-20      0       -1.31355425421E-20
      0        -9.48594322162E-24      0        6.56361379837E-24
      0         5.68778402700E-27      0       -3.93528208757E-27
      0        -3.97840565974E-30      0        2.75240356386E-30
      0         3.17997410062E-33      0       -2.19986945318E-33
      0        -2.85921776493E-36      0        1.97783856983E-36

The micro-minimality of the improvement of approximation becomes possibly even more stunning when comparing the improvement from $2*10^3, 2*10^4, 2*10^5 $ to $2*10^6$ coefficients. this gives this table:

  reciprocal      reciprocal
even convergent  odd convergent     at index    at index 
-----------------------------------------------------------
 0.908014406855  0.810882478513        2 000       2 001
 0.907973742909  0.810929315943       20 000      20 001
 0.907969675369  0.810934002994      200 000     200 001
 0.907969268604  0.810934471732    2 000 000   2 000 001

It really seems as if the approximates at even and odd convergents get stuck at separate values.(The reciprocal convergents should both approximate $\pi^3/36 \approx 0.8613$)