I am currently learning about power series solutions for second order linear ODE's. I am reading a proof of the following statement:
Let $x_0$ be an ordinary point of the differential equation $y''+p(x)y'+q(x)y=0$, (11) and let $a_0$ and $a_1$ be arbitrary constants. Then there exists a unique function $y(x)$ that is analytic at $x_0$, is a solution of equation (11) in a certain neighborhood of this point, and satisfies the initial conditions $y(x_0) = a_0$ and $y'(x_0) = a_1$. Furthermore, if the power series expansions of $P(x)$ and $Q(x)$ are valid on an interval $|x − x0|< R$, $R > 0$, then the power series expansion of this solution is also valid on the same interval
("Differential Equations With Applications And Historical Notes", by George F Simmons), page 246
In the proof, at one point he defines $r$, as some constant smaller than $R$. In the proof, for the sake of simplicity, $x_0$ is assumed to be zero, which means that we know $p(x),q(x)$ converge for $x<|R|$, and consequently, for $|x|<r$. We then let ${p_i},{q_i}$ be the sequences for the power series approximating $P(x)$,$Q(x)$. We know that on this interval, the power series exist, and therefore, converge. By the divergence test, the terms of the sequences must then approach $0$. Now, he says, there must exist some constant $M$ so:
$|p_n|*r^n<M, |q_n|*r^n<M $
"for all $n$". And this is where i get very confused. Because how do we know $|p_n|*r^n$ (or q) converges? It would happen if $|r|<1$, but that has not been given as a requirement. We can't use l'hopitals as $\frac{|p_n|}{\frac{1}{r^n}}$ either since we would have to take the derivative of $|p_n|$ which is a discrete thing. Now, it might be the case that with the "for all $n$", he simply means: "for each $n$, there exist that $M$ with that property", instead of "there exist a $M$ with that properties for that expression no matter what $n$ we plug in. But that second option i think - am not sure - might not be the case as he latter in the proof, for expressions with different values of $n$ uses the same $M$.
So my question is: does $|p_n|*r^n$ really converge (and is therefore bounded) ? And if so, why? The full proof is in the book so i can't quite link to it
The series $\sum_{n=0}^\infty p_n x^n$ converges absolutely for $\lvert x\rvert<R$, in particular for $x=r$. Let $M=\sum_{n=0}^\infty \lvert p_n \rvert r^n$. Clearly, $\lvert p_n\rvert\cdot r^n\leq M$. The same can be done for the series with coefficients $q_n$.
Note that one can even chose the constant $M$ uniformly in $r$ on compact subsets of $[0,R)$.