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