This is part of the exercise $8$ in page $216$ of Analysis I of Amann and Escher.
Show, by example, that if $\rho_a,\rho_b>0$ is possible that $\rho_{ab}>\max\{\rho_a,\rho_b\}$, were $\rho_c$ is the radius of convergence of a power series $c:=\sum c_k X^k$.
In other words: I must found an example where if I define $a:=\sum a_k X^k$, and $b:=\sum b_k X^k$ and $c:=\sum c_k X^k$ with
$$\left(\sum a_k X^k\right)\left(\sum b_k X^k\right)=\sum c_k X^k$$
and $c_k=\sum_{k=0}^n a_kb_{n-k}$, then
$$\rho_c>\max\{\rho_a,\rho_b\}$$
I tried a lot of combinations but I found nothing. In particular if we define $a_k=\alpha ^k$ and $b_k=\beta ^k$ for any $\alpha,\beta\in\Bbb R$ we have that
$$\rho_c=\max\{\rho_a,\rho_b\}$$
Do you know some elementary example for this exercise?
Let me demystify the excellent example given by Daniel Fischer to the level of pure calculus. Let $(a_n)$ and $(b_n)$ be defined by
$$ a_n = \begin{cases} 1, & n = 0 \\ 2, & n \geq 1 \end{cases} \qquad \text{and} \qquad b_n = (-1)^n a_n. $$
It is easy to check that $\rho_a = \rho_b = 1$. On the other hand, $c_0 = a_0 b_0 = 1$ and for $n \geq 1$ we have
\begin{align*} c_n &= \sum_{k=0}^{n} a_k b_{n-k} \\ &= \color{red}{(-1)^n \cdot 2} + \color{blue}{(-1)^{n-1} \cdot 4} + \cdots + \color{red}{(-1) \cdot 4} + \color{blue}{2} \\ &= (-1)^n \cdot 2 - 2 + 4 \sum_{k=0}^{n-1} (-1)^k \\ &= (-1)^n \cdot 2 - 2 + 4 \cdot \frac{1 - (-1)^n}{1 - (-1)} \\ &= 0. \end{align*}
Therefore $\rho_c = \infty$ and we have $\rho_c > \max\{\rho_a, \rho_b\}$.