# Radius of Convergence of two Power Series with Coupled Coefficents

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?

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.
• What is the problem if $1/C$ is a integer? Why shouldn't the series exists in this case? Commented Aug 5, 2019 at 6:48
• If $1/C=\pm n$ with $n$ a positive integer, then $A_n$ is not invertible, and we must have $a_{n-1}=\pm b_{n-1}$ for a solution to exist. In the end, this translates into a specific value of $B$. Commented Aug 5, 2019 at 7:03
• So in order to prove that these power series converge, one has to ensure that $|| [ a_n, b_n ] ||_2 < || [ a_{n-1}, b_{n-1}] ||_2$. Is there any theorem? It somehow a generalization of the Chauchy-Hadamard theorem? Commented Aug 10, 2019 at 16:23
• Note $\color{red}{m-1}$ (not $n-1$) above. This $m$ is just a constant; it is the product that gives an $1/n!$-like decay. Yes, the Cauchy-Hadamard theorem can be applied, but it's easier to use the ratio test (and the comparison test). Commented Aug 10, 2019 at 17:37