Let $\sum_{n=0}^{\infty}a_{n}(z-a)^{n}$ and $\sum_{n=0}^{\infty}b_{n}(z-a)^{n}$ be two power series with radii of convergence $R_{1}$ and $R_{2}$ respectively. Then the Cauchy Product of these series can be defined as $\sum_{n=0}^{\infty}c_{n}(z-a)^{n}$ where
$c_{n}= \sum_{k=0}^{n}a_{k}b_{n-k}$. Furthermore, the Cauchy product $\sum_{n=0}^{\infty}c_{n}(z-a)^{n}$ has a radius of convergence R at least larger or equal to the smaller of the two radii $R_{1}$ and $R_{2}$, that is $R≥min\{R_{1},R_{2}\}$.
Can someone explain to me why the latter is true, namely $R≥min\{R_{1},R_{2}\}$ ? I've already looked at the answer in Cauchy Product Radius of Convergence, but it doesn't become clear to me.
I know that $\sum_{n=0}^{\infty}c_{n}(z-a)^{n}=\sum_{n=0}^{\infty}a_{n}(z-a)^{n} \sum_{n=0}^{\infty}b_{n}(z-a)^{n}$ for $z \in \mathbb{C}$ with $\vert z-a \vert < min \{R_{1},R_{2} \}$, but I don't see how that might help me get further.
Best Answer
We may assume that $a=0$ without loss of generality. We know that $$ \frac{1}{R_1} = \limsup_{n\to +\infty} \sqrt[n]{|a_n|},\qquad \frac{1}{R_2}=\limsup_{n\to +\infty}\sqrt[n]{|b_n|} $$ hence for any $\varepsilon>0$ there is a $N_\varepsilon$ ensuring $$ \left|a_n\right|\leq\left(\frac{1}{R_1}+\varepsilon\right)^n,\qquad \left|b_n\right|\leq\left(\frac{1}{R_2}+\varepsilon\right)^n $$ for any $n\geq N_\varepsilon$. This leads to $$ |a_n|\leq M\left(\frac{1}{R_1}+\varepsilon\right)^n,\qquad \left|b_n\right|\leq M\left(\frac{1}{R_2}+\varepsilon\right)^n $$ for any $n\geq 0$, where the constant $M$ only depends on $\varepsilon$. By the triangle inequality
$$ |c_n| \leq \sum_{k=0}^{n}|a_k|\cdot|b_{n-k}|= M^2\sum_{k=0}^{n}\left(\frac{1}{R_1}+\varepsilon\right)^k \left(\frac{1}{R_2}+\varepsilon\right)^{n-k} $$ hence $|c_n|\leq M^2(n+1) \left(\varepsilon+\max\left(\frac{1}{R_1},\frac{1}{R_2}\right)\right)^n $ and $$ \limsup_{n\to +\infty}\sqrt[n]{|c_n|} \leq \varepsilon+\max\left(\frac{1}{R_1},\frac{1}{R_2}\right).$$ Since $\varepsilon$ is arbitrary, the radius of convergence of $\sum_{n\geq 0}c_n z^n$ is exactly the minimum between the radius of convergence of $\sum_{n\geq 0}a_n z^n$ and the radius of convergence of $\sum_{n\geq 0}b_n z^n$.