A question regarding Frobenius method in ODE Suppose $b(x),c(x)$ are real functions analytic at $0$. Let $b(x)=\sum_{i=0}^\infty b_ix^i, c(x)=\sum_{i=0}^\infty c_ix^i$ on $(-R,R)$. Suppose $r$ is a double root of  $r(r-1)+b_0r+c_0=0$. It is well known that the differential equation $$x^2y''+xb(x)y'+c(x)y=0$$ has a solution of the form $$y_1=x^r(1+\sum_{i=1}^\infty a_ix^i),$$ where the series $\sum_{i=1}^\infty a_ix^i$ has radius of convergence $\ge R$ (e.g., see Tyn Myint-U, Ordinary Differential Equations). Another solution is of the form $$y_1\ln x + x^r(\sum_{i=1}^\infty A_ix^i).$$ Most books (including Tyn's) mention without proof that $\sum_{i=1}^\infty A_ix^i$ also has radius of convergence at least $R$. 

Let $I(s)=s(s-1)+b_0s+c_0$. Then $A_i, i\ge 1$ satisfy the following recursive relation
$$I(r+i)A_i=-\sum_{k=0}^{i-1}A_k[(r+k)b_{i-k}+c_{i-k}]-2ia_i-\sum_{k=1}^{i}b_ka_{i-k}$$
where $A_0=0$. (Note that $2r=1-b_0$).

Proving that the power series $\sum_{k=1}A_kx^k$ has radius of convergence at least $R$
directly seems to be hopeless. I tried comparison test as in Coddington's book Introduction
to Ordinary Differential Equations, but with no success.

Does anyone have a proof, or a reference where a proof is given?
TCL, you are absolutely correct and the proof of Theorem 4.5 in my book is not complete. That $h_2$ has the desired radius of convergence can be shown similarly as for $h_1$; the only trick is to change the recurrence for $j\le m$ to avoid the division by $0$ at $j=m$. The details can be found in the proof of Theorem 4.13 which covers the present situation by the discussion at the end of this section on p139.

I have added this information to the errata which can be found on my webpage.
For record, I will give a direct proof here. First note that a power series $\sum_i d_ix^i$ has radius of convergence at least $R$ if and only if for any fixed positive number $t$ less than $R$, the sequence $|d_i|t^i, i\ge 0$ is bounded. (To prove, show that $\limsup |d_i|^{1/i}\le 1/t.$)

Let $s$ be such a number less than $R$. Let $B=\sum_{i=0}^\infty |b_i|s^i, C=\sum_{i=0}^\infty |c_i|s^i$. Since $\sum_{i=0}^\infty ia_ix^i$ has the same radius of convergence as $\sum_{i=0}^\infty a_ix^i$, the sequence $2ia_is^i,i=0,1,2\cdots$ is bounded, say by $D$. $\sum_{k=0}^i b_ka_{i-k}s^i$ is the $i$-th term of the product series of $\sum_{i=0}^\infty b_is^i$ and $\sum_{i=0}^\infty a_is^i$, so the sequence is bounded by a number say $E$. Choose $N$ such that $$\frac{(|r|+i)B+C}{|I(r+i)|}\le \frac{1}{2},$$ $$\frac{D}{|I(r+i)|}\le \frac{1}{4},$$ $$\frac{E}{|I(r+i)|}\le \frac{1}{4},$$ for all $i\ge N$. Choose $F\ge 1$ such that $|A_i|s^i\le F$ for $i\le N$.

Suppose $|A_j|s^j\le F$ for $j&lt;i$. Then by the recursive relation, \begin{eqnarray*}|A_i|s^i &amp;\le&amp; F\frac{(|r|+i)B+C}{|I(r+i)|}+\frac{D}{|I(r+i)|}+\frac{E}{|I(r+i)|}\\
&amp;\le&amp; \frac{F}{2}+\frac{1}{2}\\
&amp;\le&amp; F
\end{eqnarray*}
This proves by induction that $|A_i|s^i, i=0,1,2\cdots$ is bounded by $F$.