TITLE: Evaluation of a slow continued fraction
QUESTION [13 upvotes]: Puzzle question... I know how to solve it, and will post my solution if needed; but those who wish may participate in the spirit of coming up with elegant solutions rather than trying to teach me how to solve it. [paraphrased from Lone Learner]   
Evaluate the continued fraction
$$
1+\cfrac{-5}{13+\cfrac{-128}{37+\cfrac{-693}{73+\cfrac{-2240}{121+\ddots}}}}=
b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\cfrac{a_3}{b_3+\cfrac{a_4}{b_4+\ddots}}}}
$$
where $b_n = 6n^2+6n+1$, $a_n=4n^2-9n^4$.  
Can you determine whether it converges, and if so find its value?  Numerical computations suggest it converges extremely slowly.

REPLY [3 votes]: We want to apply the generating-function method with Hayman's method to determine asymptotics of the coefficients, as in 
LINK.  
Let
$$\begin{align*}
 p_n &= (6n^2+6n+1) p_{n-1}+(4n^2-9n^4) p_{n-2}, \quad p_{-1} = 1, \quad p_0=1,
 \\
 q_n &= (6n^2+6n+1) q_{n-1}+(4n^2-9n^4)q_{n-2}, \quad q_{-1} = 0, \quad q_0=1,
 \\
 r_n &= \frac{p_n}{q_n}.
\end{align*}$$
So the $r_n$ are the convergents of the continued fraction:
$$
r_0 = 1,\quad
 r_1 = 1+\cfrac{-5}{13},\quad
 r_2 = 1+\cfrac{-5}{13+\cfrac{-128}{37}},\quad
 r_3 = 1+\cfrac{-5}{13+\cfrac{-128}{37+\cfrac{-693}{73}}},
$$
Now a problem is that generating functions
$$
\sum_{n=0}^\infty p_n x^n,\qquad\text{and}\qquad
\sum_{n=0}^\infty p_n \frac{x^n}{n!}
$$
both have radius of convergence zero.  So they are no good in Hayman's method.  So we will use "Bessel" generating functions
$$
f(x) = \sum_{n=0}^\infty p_{n-1}\frac{x^n}{(n!)^2},\qquad
g(x) = \sum_{n=0}^\infty q_{n-1}\frac{x^n}{(n!)^2} .
$$  
Use the recurrences to compute the differential equations these satisfy.  They are
$$\begin{align*}
L[f]&=0,\qquad f(0)=1,\qquad f'(0)=1, \tag{1}
\\
L[g]&=1,\qquad g(0)=0,\qquad g'(0)=1, \tag{2}
\end{align*}$$
where second-order linear differential operator $L$ is given by
$$
L[\phi](x) = x(1-3x)^2 \phi''(x)+(1-3x)(1-9x)\phi'(x)+(-1+5x)\phi(x)
$$ 
Now the solution of (1) is
$$
f(x) = (1-3x)^{-1/3} = 1 + x + 2 x^2 + \frac{14}{3} x^3 + \frac{35}{3} x^4 +O(x^5) .
$$
The coefficient shows
$$
p_{n-1} \sim \frac{(n!)^2 3^n n^{-2/3}}{\Gamma(1/3)}
$$
With the help of Maple, I get the solution of (2) as
$$
g(x) = \frac{3}{2}(1-3x)^{-2/3} + 
\sqrt{3}\arctan\left(\frac{1+2(1-3x)^{1/3}}{\sqrt{3}}\right)(1-3x)^{-1/3}-
\frac{9+2\pi\sqrt{3}}{6}(1-3x)^{-1/3}
$$
and in powers of $(1-3x)$,
$$
g(x) = \frac{3}{2}(1-3x)^{-2/3}-
\left(\frac{\pi}{2\sqrt{3}}+\frac{3}{2}\right)(1-3x)^{-1/3}+\frac{3}{2}+
O\left((1-3x)^{1/3}\right) .
$$
Now we need to examine the coefficients at $x=0$ for the Taylor series.  The first term has coefficient given asymptotically by $C 3^n n^{-1/3}$. 
The second term by $C 3^n n^{-2/3}$, the constant term by $0$.  So overall we get
$$
q_{n-1} \sim C (n!)^2 3^n n^{-1/3} .
$$
So the convergent
$$
r_{n-1} = \frac{p_{n-1}}{q_{n-1}} \sim C n^{-2/3+1/3}
$$
goes to zero.  The value of the continued fraction is $0$.  
Note that the differential operator $L$ in terms of $y=1-3x$ is
$$
9y^2(1-y)\phi''(x)+9y(2-3y)\phi'(y)+(2-5y)\phi(y)
$$
with indicial polynomial $9r(r-1)+18r+2 = (3r+2)(3r+1)$ roots $-1/3, -2/3$.  This is the reason for the type of series we got.