<s>[INST] Radius of convergence of $\sum_{n=0}^\infty a_n x^n$, with $a_{n+2} = \frac{n(n+1) a_{n+1} - a_n}{(n+2)(n+1)}, a_2 = -a_0/2$ Problem

Find the radius of convergence of the power series
$$
\sum_{n=0}^\infty a_n x^n
$$

where $a_n$'s are defined by the following recurrence relation

$$
\begin{aligned}
a_{n+2} &= \frac{n(n+1) a_{n+1} - a_n}{(n+2)(n+1)}, n\ge 1 \\[8pt]
a_2 &= -a_0/2
\end{aligned}
$$

with arbitrary $a_0, a_1$.

Try1
I have tried to directly applying the ratio test, by dividing the above recurrence relation by $a_{n+1}$,
$$
r_{n+1} = \frac{n}{n+2} - \frac{1}{(n+2)(n+1)} \frac{1}{r_n}
$$
where $r_n := a_{n+1}/a_n$. We can observe that
$$
r_n r_{n+1} = \frac{n}{n+2} r_n - \frac{1}{(n+2)(n+1)}
$$
but I cannot proceed to find the expression for
$$
\lim_{n \to \infty} \vert r_n \vert
$$

Try2
Since the recurrence relation depends on the arbitrary choice pf $a_0, a_1$, let us proceed by letting $a_0:= 0$. We have
$$
\begin{aligned}
a_2 &= a_0 = 0 \\[7pt]
a_3 &= -\frac{1}{6} a_1 \\[7pt]
a_4 &= \frac{2}{4} a_3 =  -a_1/12 \\[7pt]
a_5 &= \frac{3}{5} a_4 - \frac{1}{5\cdot 4}a_3 = -\frac{1}{24} a_1 \\[7pt]
\end{aligned}
$$
where I cannot find any simple rules. Likewise, by letting $a_1:= 0$, we have
$$
\begin{aligned}
a_2 &= -a_0/2 \\[7pt]
a_3 &= \frac{1}{3} a_2 = -\frac{1}{6}a_0 \\[7pt]
a_4 &= \frac{2}{4} a_3 - \frac{1}{4\cdot 3}a_2 =  -\frac{1}{24} a_0\\[7pt]
a_5 &= \frac{3}{5} a_4 - \frac{1}{5\cdot 4}a_3 = -\frac{1}{60} a_0 \\[7pt]
\end{aligned}
$$
where again I have failed to find any rules. So, I cannot find the interval that the following composition is valid.
$$
\sum_{n=0}^\infty a_n x^n = a_0 \left[  1 - x^2/2 - x^3/6 - x^4/24 - x^5/60 + \cdots \right] + a_1 \left[ x - x^3/6 - x^4/12 - x^5/24 + \cdots\right]
$$
Any help will be appreciated.
 [/INST] The trick is to notice $a_{n+2}=-\frac{f_n}{(n+2)!}(a_0+a_1)$ for $n\ge1$ with $f_1=f_2=1,\,f_n=nf_{n-1}-f_{n-2}$. Asymptotically $f_n\sim nf_{n-1}$, so $\frac{f_n}{(n+2)!}$ is a sequence in which the ratio of consecutive terms $\to1$. Thus the radius of convergence is also $1$.
</s>