Q:

Radius of convergence of product

Let $\sum_{i=0}^\infty a_nz^n$ and $\sum_{i=0}^\infty b_nz^n$ be power series, and define the product $\sum_{i=0}^\infty c_nz^n$ by $c_n=a_0b_n+a_1b_{n-1}+\ldots+a_nb_0$. Find an example where the first two series has radius of convergence $R$, while the third (the product) has radius of convergence larger than $R$.

The radius of convergence of $\sum_{i=0}^\infty a_nz^n$ is given by $1/R=\limsup{|a_n|^{1/n}}$. I tried some sequences like $a_0=a_1=\ldots=b_0=b_1=\ldots=1$. Then the two sequences have radius $1$. But $c_i=i+1$, and $\lim_{i\rightarrow\infty}(i+1)^{1/i}=1$. So the radius is the same as the original two sequences, which doesn't work. 

A:

A simpler example: let 
$$
f(z) = \frac{1+z}{1-z} = \frac{1}{1-z} + \frac{z}{1-z}. 
$$
Note that the first term is just the formula for the geometric sum with first term 1, $$\frac{1}{1-z} = 1 + z + z^2 + z^3 + \cdots, \qquad |z| < 1,$$ and the second term is the formula for a geometric sum with first term equal to the common ratio $z$:
$$
\frac{z}{1-z} = \frac{1}{1-z} - 1 = z + z^2 + z^3 + \cdots, \qquad |z| < 1. 
$$
Then the power series for $f(z)$ is given by
$$
f(z) = \frac{1+z}{1-z} = 1 + 2z + 2z^2 + 2z^3 + \cdots = 1 + 2\sum_{n=1}^\infty z^n, \qquad |z| < 1,
$$
and has radius of convergence $R_f = 1$.
If we form a new power series $g(z)$ by making the substitution $z \mapsto -z$, we have
$$
g(z) = \frac{1-z}{1+z} = 1 - 2z + 2z^2 - 2z^3 + \cdots = 1 + 2\sum_{n=1}^\infty (-z)^n, \qquad |z| < 1,
$$
also with radius of convergence $R_g = 1$. However, the product series is
$$
f(z)g(z) = \left( \frac{1+z}{1-z} \right) \left( \frac{1-z}{1+z} \right) = 1 = 1 + 0z + 0z^2 + 0z^3 + \cdots, \qquad \forall z\in\mathbb{C}
$$
and has radius of convergence $R_{fg} = \infty$, which is strictly larger than $R_f = R_g = 1$.

A:

Let $f(x)=(1-x)^{1/2}$ and $g(x)=(1-x)^{-1/2}$.  When expanded in a Maclaurin series, we get two series with radius of convergence $1$. 
The Cauchy product (your product) of the two series is the very simple "infinite" series $1+0\cdot x+0\cdot x^2+\cdots$, which has infinite radius of convergence. 
Remark: If the example is too simple, we can "doctor" $f(x)$ by mutiplying say it by $h(x)$, where $h(x)=\frac{1}{1-\frac{x}{3}}$. Then the Cauchy product  of the Maclaurin series for $f(x)h(x)$ and $g(x)$ has radius of convergence $3$. 

