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I need help with finding the radius of the Cauchy product of two aboslutely convergent series.

Concretely, let $\sum a_k z^k $ have radius of convergence equal to $R_1$ and $\sum b_k z^k $ have radius of convergence equal to $R_2$.

Let $\sum c_k z^k$ be the Cauchy product of the two series.

I am stuck trying to calculate the radius of convergence of $\sum c_k z^k$.

Please could someone show me how to calculate the radius of convergence of the Cauchy product of two absolutely convergent series?

The correct result is $R = \min (R_1, R_2)$ as the exercise asks to show that the Cauchy product converges for $|z|<\min(R_1, R_2)$.

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  • $\begingroup$ The way you formulated it, your question doesn't make any sense: $\sum a_k$ is a series of (complex) numbers, not depending on any variable. Such an object doesn't have a radius of convergence, either it is convergent or not! A more appropriate formulation would be: $\sum \limits_{k = 0}^\infty a_k (z - z_0)^k$ with $z_0 \in \mathbb{C}$. $\endgroup$
    – ComplexF
    Sep 22, 2016 at 14:17
  • $\begingroup$ You're right. I corrected it. $\endgroup$
    – student
    Sep 22, 2016 at 21:18

3 Answers 3

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I assume that you are talking of two power series.

By Mertens Theorem, the radius of convergence of the Cauchy product is at least the minimum of $R_1$ and $R_2$. However it can be bigger than that. For example take $A(x)$ and $B(x)$ be respectively the Taylor series centred at $0$ of $\sqrt{1-x}$ and $\frac{1}{\sqrt{1-x}}$ then $R_1=R_2=1$, whereas $C(x)=1$ has an infinite radius of convergence.

For more detail see Radius of convergence of product

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There's no general solution, it should depend on the series. But assuming both series are centered at the same point, $R= \min\{R_1,R_2\}$ is a lower bound for the radius of convergence. I'll let you figure out the adjustment if they're centered at different points.

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You know that A (resp. B) for each $z \in \mathbb{C}$ with $|z| < R_1$ (resp. $|z| < R_2$) converges.

Moreover you should know that (1) the cauchy product of two absolute convergent series is also absolute convergent and (2) if a power series converges for all $z$ with $|z| < r$ than the radius of convergences is greater equal than r.

Let $z \in \mathbb{C} < min\{R_1,R_2\}$ be arbitray. Than A and B convergence absolutely and therefore the Cauchy Product converges absolutely by (1). Therefore $r \geq \min\{R_1,R_2\}$

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