# Cauchy multiplaction

Would like to know if exists an example for $$\sum_0^\infty a_n x^n,\sum_0^\infty b_n x^n$$ $$\sum_0^\infty c_n x^n, c_n:=\sum_{k=0}^n a_k b_{n-k}$$ such that $\max\{R_a,R_b\} < R_c < \infty$ ($R$ stands for convergence radius)

??

(Couldn't find any reason there shouldn't be, but can't find an example)

Thanks!

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So there is some $\hat{x}$ where neither $A(\hat{x})$ nor $B(\hat{x})$ converges, but $A(\hat{x})B(\hat{x})$ does? – vadim123 May 8 '13 at 20:33

Try $f(x)g(x)=\frac1{1-x}$, $f(x)=\frac{1-3x}{1-2x}$, and hence of course $g(x)=\frac{1-2x}{(1-x)(1-3x)}$. Because of the poles, $f$ will have $R_a=\frac12$, $g$ will have $R_b=\frac13$ and $fg$ will have $R_c=1$.

To give the explicit coefficients:

• $f(x)=\sum a_nx^n$ with $a_n=-2^{n-1}$ for $n>0$ and $a_0=1$.
• $g(x)=\sum b_nx^n$ with $b_n=\frac{5+\sqrt 5}{10}(\frac{3+\sqrt 5}2)^n + \frac{5-\sqrt 5}{10}(\frac{3-\sqrt 5}2)^n$ (from the recursion $b_{n+1}=3b_n-1$).
• $c_n=\sum_{k=0}^na_kb_{n-k}$ will turn out to be $1$.
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