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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.

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  • $\begingroup$ Are you familiar with the Mertens theorem? $\endgroup$ – Demophilus Jun 9 '17 at 13:01
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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 $\sum_{n\geq 0}b_n z^n$.

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  • $\begingroup$ Hi, can you explain how you obtained the constant $M$? $\endgroup$ – Nicholas Roberts Feb 8 at 16:42
  • $\begingroup$ @NicholasRoberts: as the maximum between $\sup_{n\leq N_\varepsilon}|a_n|\left(\frac{1}{R_1}+\varepsilon\right)^{-n}$ and $\sup_{n\leq N_\varepsilon}|b_n|\left(\frac{1}{R_2}+\varepsilon\right)^{-n}$. $\endgroup$ – Jack D'Aurizio Feb 8 at 19:14
  • $\begingroup$ It's clear to me how the estimate $|a_n|\leq M\left(\frac{1}{R_1}+\varepsilon\right)^n$ work for $n \leq N$. But I dont see how it works for $|a_n|$ where $n > N$. I know for such $n$ we have that $\left|a_n\right|\leq\left(\frac{1}{R_1}+\varepsilon\right)^n$. But how do we introduce $M$ into this? $\endgroup$ – Nicholas Roberts Feb 9 at 18:06
  • $\begingroup$ @NicholasRoberts: you have an inequality that works for any $n$ sufficiently large. I am just stating that by weakening it by a constant factor you get a similar inequality that works for any $n$. $\endgroup$ – Jack D'Aurizio Feb 9 at 19:53
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Basic fact: The radius of convergence of $\sum_{n=0}^{\infty}c_{n}(z-a)^{n}$ is the unique $R\in [0,\infty)$ such that $\sum c_n(z-a)^n$ converges for $|z-a|<R,$ and diverges for for $|z-a|>R.$

Now you wrote this: "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." Thus you appear to know that $\sum c_n(z-a)^n$ converges if $\vert z-a \vert < \min \{R_{1},R_{2} \}.$ Now think about it: Given the basic fact above, isn't it clear we must have $R\ge \min \{R_{1},R_{2} \}?$

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