Convergence of a power series function Consider the following differential equation:
$$w''(x)+p(x)w'(x)+q(x)w(x)=r(x)$$
with the initial condition of $w(0)=w_0,\ w'(0)=w_1$, and
$$w_{n+2}=\frac{r_{n+2}-(n+1)p_0w_{n+1}-\sum_{k=0}^n w_k (q_{n-k}+kp_{n+1-k})}{(n+1)(n+2)}$$
where $p$, $q$ and $r$ possess power-series expansions of the form $y(x)=\sum_{n=0}^\infty y_n x^n$,
with $\{p_n\},\{q_n\},\{r_n\}$ respectively, are analytic and converge in the interval $(-R,R)$.
$\forall w_0,w_1$ the aforementioned differential equation, has a unique solution in $(-R,R)$, $ w(x)$, which is analytic and has a power-series expansion with $\{w_n\}$.
We want to show that $w(x)=\sum_{n=0}^\infty w_n x^n$ converges in the interval as well.
 A: OK, I will put an answer here.  This is the kind of thing that someone who wants/has to do analysis should learn to do themselves.  And the way to learn it is by practice.
Let $x_0$ be such that $|x_0|<R$.  Choose $T$ with $|x_0|<T<R$.  We will show that sequence $|w_n| T^n$ is bounded.  From this (for example by the root test) we conclude that power series $\sum w_nx^n$ has radius of convergence $\ge T$, and so in particular the series converges at $x_0$.
Since $T$ is within the radius of convergence of series $\sum r_n x^n$, we have $r_n T^n \to 0$, so it is bounded; that is, there is a constant $A$ such that
$$
|r_n| T^n \le A\qquad\text{for all $n$}.
$$
Similarly, there is a constant $C$ such that
$$
|q_n|T^n \le C \qquad\text{for all $n$}.
$$
For the $p$ series we need more.  Since $|p_n| T^n \to 0$, also the weighted average
$$
\frac{2}{n(n+1)}\sum_{k=0}^n k|p_{n+1-k}|T^{n+1-k} \to 0
$$
Now, $p_0T/(n+2) \to 0$ as $n \to \infty$ and $(T^2C)/(n+2) \to 0$ as $n \to \infty$, so there exists $N \in \mathbb N$ so that
$$
\frac{|p_0|T}{n+2} \le \frac{1}{4},\qquad
\frac{T^2C}{n+2} \le \frac{1}{4},\qquad
\frac{T}{(n+1)(n+2)}\sum_{k=0}^n k|p_{n+1-k}|T^{n+1-k} \le \frac{1}{4}
$$
for all $n \ge N$.  (Where did these come from?  They were chosen so that the argument below works.  But in an orgainized proof, they are placed here, so the reader can see what they depend on.)  Now $K$ will be the maximum of a finite list of values:
$$
4A,\quad
|w_0|, |w_1| T, |w_2|T^2, |w_3| T^2,\cdots, |w_N|T^N, |w_{N+1}|T^{N+1} .
$$
Now I claim that $|w_n| T^n \le K$ for all $n \in \mathbb N$.  The proof is by induction.  It is true for $0 \le n \le N+1$ by the definition of $K$.  Now suppose it is true up to some $n+1$ and consider $n+2$.  Plug in the recurrence to get
$$
w_{n+2}T^{n+2} = \frac{r_{n+2}T^{n+2}-(n+1)p_0 T w_{n+1}T^{n+1}
-\sum_{k=0}^n w_kT^k(T^2q_{n-k}T^{n-k}+kTp_{n+1-k}T^{n+1-k})}{(n+1)(n+2)}
$$
But since $n > N$
$$\begin{align}
\frac{|r_{n+2}|T^{n+2}}{(n+1)(n+2)} &\le\frac{A}{(n+1)(n+2)} \le A \le \frac{1}{4}\;K
\\
\frac{(n+1)|p_0| T |w_{n+1}|T^{n+1}}{(n+1)(n+2)} &\le
\frac{|p_0| T K}{(n+2)} \le \frac{1}{4}\;K
\\
\frac{\sum_{k=0}^n |w_k|T^k T^2|q_{n-k}|T^{n-k}}{(n+1)(n+2)} &\le \frac{(n+1)}{(n+1)(n+2)}K T^2 C \le \frac{1}{4}\;K
\\
\frac{\sum_{k=0}^n |w_k|T^k kT|p_{n+1-k}|T^{n+1-k}}{(n+1)(n+2)} &\le
\frac{KT}{(n+1)(n+2)}\sum_{k=0}^n k|p_{n+1-k}|T^{n+1-k}
\le  \frac{1}{4}\;K
\end{align}$$
and thus $|w_{n+1}|T^{n+2} \le 4 K/4 =K$.  This completes the induction.
