Radius of convergence of Cauchy product – examples

Question

Assume two power series $\sum_{n\ge0}a_n x^n=f_a(x),\sum_{n\ge0}b_n x^n=f_b(x)$ with radii $r_a,r_b$ (respectively) and $r_a\lneqq r_b.$ Consider their Cauchy product $$\sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k}\right)x^n$$ and its radius of convergence $r$. We know that $r\ge r_a$. Now, could you give an example of functions such that $f_a(x)f_b(x)$ has a radius $r_a\lneqq r\lneqq r_b$? And is it possible to have $r_b\lneqq r\lneqq\infty$ (if so, could you give an example again)?

Answer 1

Consider the functions

$$ f(x) = \frac{1}{1-x} \frac{1}{1- x/2}, \qquad g(x) = (1-x),$$

without computing the power series representations it is clear that $f$ has radius of convergence $R_f = 1$, whilst $g$ has radius of convergence $R_g = \infty$.

However, the product

$$h(x) = f(x)g(x) = \frac{1}{1-x/2},$$

has radius of convergence $R_h = 2$, and in particular $R_f < R_h < R_g$.

Now if we consider instead

$$ f(x) = \frac{1}{\sqrt{1-x} } \frac{1}{1-x/2}, \qquad g(x) = \sqrt{1-x},$$

then we now have $R_f = R_g = 1$ and as before defining $h(x) = f(x)g(x)$ we have $R_h = 2$, so that now $R_f = R_g < R_h$.