Radius of Convergence of the Power Series Solution to a Second Order Linear Homogeneous ODE I was studying the proof of the following theorem 

Theorem. Let $x_0$ be a real number and suppose that the coefficients $a(x)$, $b(x)$ in
  $$L[y](x)=y^{''}(x)+a(x)y^{'}(x)+b(x)y(x)\tag{1}$$
  have convergent power series expansions in powers of $x-x_0$ in some interval as follows
  $$a(x)=\sum_{n=0}^{\infty}a_n(x-x_0)^n, \quad b(x)=\sum_{n=0}^{\infty}b_n(x-x_0)^n, \quad |x-x_0| \lt r_0, \quad r_0 \gt 0\tag{2}$$
  Then there exists a nontrivial solution $\phi$ of the problem
  $$L[y](x)=0 \tag{3}$$
  with a power series expansion
  $$\phi(x)=\sum_{n=0}^{\infty}c_n(x-x_0)^n\tag{4}$$
  convergent for $|x-x_0|<\rho$ where $\rho \ge r_0$.

Proof Outline
The proof is straight forward. It considers the power series expansions of the coefficients in $(2)$ which are convergent for $|x-x_0| \lt r_0,\,r_0 \gt 0$. Then it puts the series solution $(4)$ into the ODE mentioned in $(3)$ and obtains a recurrence relation for the $c_i$s such that the ODE is satisfied. Then it proves that the series solution $(4)$ whose coefficients obey the aforementioned recurrence relation will converge in the interval $|x-x_0| \lt r_0$. 
The following links show the original proof if you want to check it out.
Proof, Page 1
Proof, Page 2
Proof, Page 3

Question
According to the proof, it can be concluded that the solution will converge for $\rho=r_0$. 
The proof does not mention anything about the last sentence of the theorem that $\rho \ge r_0$ which is saying that $r_0$ is lower bound for radius of convergence of the solution.
I just cannot understand that how the radius of convergence of the solution can be $\rho \gt r_0$. I think having $\rho>r_0$ is meaningless as the coefficients of ODE can just be replaced by their power series only in $|x-x_0|<r_0$.
Can someone shed some light on this?
 A: I think, I have found an example for you. Suppose $a(x)=\frac{1}{1+x^{2}}$ and $b(x)=-1-\frac{1}{1+x^{2}}$. Then, obviously, both power series have radius of convergence $1$ with $x_{0}=0$ but as solution you can find
\begin{equation}
y(x)=\exp(x).
\end{equation}
So there is at least one solution which has greater radius of convergence, (even infinity) and for that both $a(x)$ and $b(x)$ can still have the same radius of convergence.
A: The followings are other examples that show there may exist nontrivial solutions which are analytic at $x=0$ with their radius of convergence being greater than that of coefficients of the ODE.
$$\begin{array}{rcll}
y''+ \dfrac{x}{1-x} y'-\dfrac{1}{1-x} y=0, & & y=x \\
y''- 2\dfrac{x}{1-x^2} y'+ \dfrac{n(n+1)}{1-x^2} y=0, & & y=P_n(x), & \text{Legendre Differential Equation}
\end{array}$$
There is also a nice example in the answer by @Alex. However, in all of the examples, the radius of convergence of the solution we are interested in is $\infty$. I couldn't find some example which the radius of convergence of solution is greater than that of coefficients but it is not $\infty$. 
I think any arguments about $ρ>r_0$, in such a proof, is meaningless as the coefficients of ODE can just be replaced by their power series only in $|x−x_0|<r_0$. But sometimes, luckily, the solution we obtained formally by this method satisfies the ODE while we have not replaced the coefficients by their power series. In this case, what just matters is the radius of convergence of the solution itself.
