Step 1: Deriving the recurrence.

Substituting $y = x^r \sum_{n=0}^{\infty} a_n x^n$ into $x^2 y'' + xp(x)y' + q(x)y = 0$ and collecting the coefficient of $x^{r+n}$:

$I(r+n) a_n + \sum_{k=0}^{n-1} [(r+k)p_{n-k} + q_{n-k}] a_k = 0,$

where $I(s) = s(s-1) + p_0 s + q_0$ is the indicial polynomial. For $r = r_1$ and $n \geq 1$:

$a_n = -\frac{1}{I(r_1 + n)} \sum_{k=0}^{n-1} [(r_1+k)p_{n-k} + q_{n-k}] a_k.$

Since $I(r_1 + n) = (r_1 + n - r_1)(r_1 + n - r_2) + \text{linear terms} = n(n + r_1 - r_2)$ (up to the leading behavior), and $r_1 - r_2$ is not a negative integer, we have $I(r_1 + n) \neq 0$ for all $n \geq 1$.

Step 2: Bounding the indicial factor.

For $n \geq 1$: $|I(r_1 + n)| \geq |n(n + r_1 - r_2)| - O(n) \cdot |p_0 - 1|$. For large $n$, $|I(r_1 + n)| \geq cn^2$ for some constant $c > 0$. Choose $N_0$ such that $|I(r_1 + n)| \geq n$ for all $n \geq N_0$.

Step 3: Majorant construction.

Since $p$ and $q$ converge for $|x| < R$, choose $\rho < R$ and let:

$|p_n| \leq \frac{P}{\rho^n}, \quad |q_n| \leq \frac{Q}{\rho^n}$

for constants $P, Q > 0$. Define $M = \max(P(\max_{0 \leq k \leq N_0}|r_1 + k|) + Q, \ldots)$ and let $K = M/\rho$.

The recurrence gives:

$|a_n| \leq \frac{1}{|I(r_1+n)|} \sum_{k=0}^{n-1} \frac{(|r_1+k|P + Q)}{\rho^{n-k}} |a_k|.$

For $n \geq N_0$, using $|I(r_1+n)| \geq n$:

$|a_n| \leq \frac{C}{n\rho^n} \sum_{k=0}^{n-1} \rho^k |a_k|$

for a constant $C$ depending on $r_1, P, Q$.

Step 4: Comparison with a geometric series.

Define $A_n = \rho^n |a_n|$. Then:

$A_n \leq \frac{C}{n} \sum_{k=0}^{n-1} A_k.$

Let $S_n = \sum_{k=0}^{n} A_k$. Then $A_n \leq (C/n) S_{n-1}$ and $S_n = S_{n-1} + A_n \leq S_{n-1}(1 + C/n)$.

By induction: $S_n \leq S_0 \prod_{k=1}^{n}(1 + C/k)$. Since $\prod(1 + C/k) \sim n^C$ (by taking logarithms: $\sum \ln(1+C/k) \sim C\ln n$), we get $S_n = O(n^C)$, hence $A_n = O(n^{C-1})$ and:

$|a_n| \leq \frac{n^{C-1}}{\rho^n}.$

Therefore $\sum |a_n| |x|^n \leq \sum n^{C-1} (|x|/\rho)^n$, which converges for $|x| < \rho$. Since $\rho < R$ was arbitrary, the series converges for $|x| < R$. $\blacksquare$