Radius of Convergence of two Power Series with Coupled Coefficents

Question

I have two functions $P(r)$ and $Q(r)$ that can be expressed in the following power series

$$P(r) = \sum^\infty_{n=0} a_n r^n$$

$$Q(r) = \sum^\infty_{n=0} b_n r^n$$

where $r \in \mathbb R_{\ge 0}$.

The coefficents $a_n$ and $b_n$ are defined by the following recurrence relation

$$a_n = C\, n b_n - a_{n-1}$$

$$b_n = C\, n a_n + b_{n-1}$$

$$a_0 = B \,b_0$$

where $a_n, b_n, C,B \in \mathbb R$.

This recurrence relation stems from inserting the power series into a set of coupled ordinary differential equations. I performed some numerical experiments already to determine the rate and radius of convergence. However, it would be nice to have some analytical expression here. Sadly, I have great difficulties applying common convergence criterion (e.g. root test, ratio test) here because of the interdependence of the coefficients.

Is there any way to get at least an approximation for the radius of convergence?

Answer 1

If $C\neq 0$ (otherwise the result is trivial) and the solution exists (it surely does if $1/C$ is not an integer; otherwise there is a condition on $B$ which is not hard to get from what follows), then the answer is $\color{red}{\infty}$, i.e. both series converge everywhere. Let's prove it. Writing $$\begin{bmatrix}a_{n-1}\\b_{n-1}\end{bmatrix}=A_n\begin{bmatrix}a_n\\b_n\end{bmatrix}, \qquad A_n=\begin{bmatrix}-1&Cn\\-Cn&1\end{bmatrix}$$ we see that, for a positive integer $m$ such that $|C|m>1$, and $n\geqslant m$, $$\begin{bmatrix}a_n\\b_n\end{bmatrix}=\left(\prod_{k=m}^{n}A_k\right)^{-1}\begin{bmatrix}a_{m-1}\\b_{m-1}\end{bmatrix}.$$ Now (again for $n\geqslant m$) the norm $\lVert A_n^{-1}\lVert_2$ is easy to compute: the matrix $$(A_n^{-1})^T A_n^{-1}=\frac{1}{(C^2 n^2-1)^2}\begin{bmatrix}C^2 n^2+1 & -2Cn \\ -2Cn & C^2 n^2 + 1\end{bmatrix}$$ has eigenvalues $(Cn\pm 1)^{-2}$, which implies $\lVert A_n^{-1}\lVert_2=1/(|C|n-1)$ and gives $$a_n^2+b_n^2\leqslant(a_{m-1}^2+b_{m-1}^2)\prod_{k=m}^{n}(|C|k-1)^{-2}.$$ The claim follows easily.