On expressing $\frac{\pi^n}{4\cdot 3^{n-1}}$ as a continued fraction.
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.)
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 use200 000
and200 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 $\pi^3/36 \approx 0.8613$)