If the radius of convergence of the power series $a$ is positive then it "reciprocal" power series have positive radius of convergence

Question

Im trying to solve this exercise from the book Analysis I of Amann and Escher (page 216, exercise 9).

Let $a=\sum a_k X^k\in\Bbb C[\![X]\!]$ with $a_0=1$.

(a) Show that there is some $b\in\Bbb C[\![X]\!]$ such that $ab=1$. Provide a recursive algorithm for calculating the coefficients $b_k$.

(b) Show that the radius of convergence of $\rho_b$ of $b$ is positive if $\rho_a$ of $a$ is positive.

The first part is easy, we have that $b_0=1$ and

$$b_n=-a_n-\sum_{k=1}^{n-1}a_kb_{n-k},\quad\forall n\ge 1$$

(with the convention that the empty sum is zero). But Im stuck in the second part. To context the exercise: this exercise comes prior to any definition of continuity, derivative or analyticity in the book, then, from this context, I dont know how to prove it or if it is provable.

My work so far: let $a:=\sum a_k X^k$ a formal power series with radius of convergence $\rho_a>0$, then for $a(x):=\sum_{k=0}^\infty a_k x^k$ for $x\in\Bbb B(0,\rho_a)$ we have

$$a_0=1\le\left|\sum_{k=0}^\infty a_kx^k\right|\le \sum_{k=0}^\infty |a_k|r^k=M<\infty,\quad\forall x\in\Bbb B(0,r),\text{ with }0

If we define $b:=\sum b_k X^k$ such that $ab=1$ then from above we have that

$$\frac1M\le\left|\sum_{k=0}^\infty b_kx^k\right|\le 1,\quad\forall x\in\Bbb B(0,r),\text{ with }0

Some help will be appreciated, thank you.

Answer 1

I found a solution that dosn't use continuity theorems, just basic properties of convergent sequences, so it fit well to the context where it is asked.

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(a) If $ab=1$ it must be the case that $a_0=b_0=1$ and $$ c_n:=\sum_{k=0}^n a_k b_{n-k}=0,\quad\forall k\in\Bbb N_{\ge 1}\tag{1} $$ From (1) we have that $$ 0=\sum_{k=1}^{n-1}a_kb_{n-k}+b_na_0+a_nb_0\implies b_n=-a_n-\sum_{k=1}^{n-1}a_kb_{n-k}\tag{2} $$ From (2) is clear that the coefficients $b_n$ are well defined, hence $b$ exists for all $a$ such that $a_0=1$.

(b) Let $a:=\sum a_k X^k$ a formal power series with radius of convergence $\rho_a>0$, then for $\underline a(x):=\sum_{k=0}^\infty a_k x^k$ for $x\in\rho_a\Bbb B$ we have $$ \left|\sum_{k=0}^\infty a_kx^k\right|\le \sum_{k=0}^\infty |a_k|r^k=M<\infty,\quad\forall x\in r\Bbb B,\text{ with }00\tag7 $$ So we only need to prove that such $\epsilon>0$ exists. Set $K:=\max_{k\in\Bbb N_{\ge0}}|a_k|$, then we have the bound $$ \sum_{k=1}^\infty|a_k| x^k\le K\sum_{k=1}^\infty x^k=\frac{Kx}{1-x},\quad\text{for }x\in(0,1)\tag8 $$ Then for any $\epsilon\in(0,\frac1{K+1} \wedge r)$ we find that $$ \sum_{k=1}^\infty|a_k|\epsilon^k\le K\frac{\epsilon}{1-\epsilon}\le 1\tag9 $$ as desired.