# Radius of convergence of product of two absolutely convergent power series

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 $|z|<\min(R_1, R_2)$.

• 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 $z_0 \in \mathbb{C}$. Sep 22, 2016 at 14:17
• You're right. I corrected it. Sep 22, 2016 at 21:18

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

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.

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 $r \geq \min\{R_1,R_2\}$