Series of sequence converges? Given the recursively defined sequence
$$
a_2 = 2(C+1)a_0
$$
and
$$
a_{n+2}=\frac{\left(n(n+6)+4(C+1)\right)a_n - 8(n-2)a_{n-2}}{(n+1)(n+2)}.
$$
that I got from the Frobenius method applied to an ODE, where for every odd $n$ we have $a_n=0$.
This brought me to the following problem:
For which $C\in \mathbb{R}$ do we have that $|\sum_{n=0}^{\infty}a_n |< \infty$?
 Of course, there is a trivial solution $C=-1$, but are there more available?
Edit: Furthermore, it was explained in the comments below that the convergence (due to linearity) is independent of the initial value $a_0$, so we could also just set this one $=1$.
If anything is unclear, please let me know. Since nobody answered so far, I just wanted to say that also interesting observations are highly appreciated. 
 A: Just an observation for now.
If we set $b_k = a_{2k}$ then we have:
$$\frac{b_{k+1}}{b_k}=\left(1+\frac{10-10k-4C}{(2k+1)(2k+2)}\right)+\frac{8(2k-2)}{(2k+1)(2k+2)}\left(1-\frac{b_{k-1}}{b_k}\right),\tag{1}$$
where:
$$\left(1+\frac{10-10k-4C}{(2k+1)(2k+2)}\right)=1-\frac{5}{2k}+O\left(\frac{1}{k^2}\right),\quad \frac{8(2k-2)}{(2k+1)(2k+2)}=\frac{4}{k}+O\left(\frac{1}{k^2}\right),$$
so assuming that $\frac{b_{k+1}}{b_k}$ behaves like $1-\frac{L}{k}$, it follows that $L=\frac{5}{2}$, hence $b_n\ll n^{-5/2}$ grants that $\sum a_n$ is convergent.

Moreover, if we set $h(x)=e^{4x^2}\sqrt{1-x^2}$ we have that the original ODE
$$(1-x^2)\,f''+x\,(7-8x^2)\,f'+4(C+1)\,f=0\tag{2}$$
is mapped into:
$$\frac{d}{dx}\left(h\cdot f'\right) = -4(C+1)\frac{h\cdot f}{1-x^2}.\tag{3}$$
Maybe the last equation is easier to solve by decomposing $h$ and $f$ with respect to a base of orthogonal polynomials (Legendre polynomials or Chebyshev polynomials, having a well-known behaviour in $1$). It is worth noticing that the Fourier series of $h$ can be expressed in terms of values of a Bessel function, that are pretty fast decaying, hence approximated solutions should be not too difficult to write.
