# If the radius of convergence of the power series $a$ is positive then it “reciprocal” power series have positive radius of convergence

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

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

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

$$\frac1M\le\left|\sum_{k=0}^\infty b_kx^k\right|\le 1,\quad\forall x\in\Bbb B(0,r),\text{ with }0<r<\rho_a$$ but from the last expression I cannot conclude that $b$ is absolutely convergent for all $|x|<\rho_b$ for some $\rho_b$, hence I cannot conclude that $b$ have a positive radius of convergence.

Some help will be appreciated, thank you.

• I found a complete solution here, sadly it imply some knowledge on continuity of series in it radius of convergence. – Masacroso Feb 1 '17 at 9:58

(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 $$\left|\sum_{k=0}^\infty a_kx^k\right|\le \sum_{k=0}^\infty |a_k|r^k=M<\infty,\quad\forall x\in r\Bbb B,\text{ with }0<r<\rho_a\tag3$$ Then suppose that there is an $\epsilon\in(0,r)$ such that $$\sum_{k=1}^\infty |a_k|\epsilon^k\le1\tag4$$ From $(2)$ we can write the bound $$|b_n|\le \sum_{k=1}^n|a_kb_{n-k}|\tag5$$ Now from (4) and (5) we want to prove by induction that $|b_n|\le \epsilon^{-n}$. Observe that $b_0=1$ so the base case holds. Now assume that $|b_n|\le \epsilon^{-n}$ and from $(5)$ we find that $$|b_{n+1}|\le \sum_{k=1}^{n+1}|a_k|\epsilon^{-n-1+k}\le \epsilon^{-n-1}\tag6$$ where the second inequality is a consequence of $(4)$. Thus by Hadamard formula we find that $$\rho_b=\frac1{\limsup\sqrt[n]{|b_n|}}\ge \epsilon>0\tag7$$ 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.