# Convergence radius of the product strictly bigger than the minimum.

Given two complex series $$f(z) := \sum z^n$$ and $$g(z) := 1 - z$$, is it true and if so, how can we show that the absolute convergence radius of their Cauchy product $$r(fg)$$ is strictly bigger than the $$\min\{r(f), r(g)\}$$?

The absolute convergence radius $$r(f)$$ must be 1 while the convergence radius $$r(g)$$ must be infinity because the sum is well-defined regardless of the value of $$z$$.

I just can't see why their product must have the radius strictly bigger than the above $$\min$$.

In this case you can compute the coefficients of the cauchy product explicitly: since $$f(z) = \sum a_k z^k$$ where $$a_k = 1$$ for all $$k=0, 1, 2, ...$$ and $$g(z) = \sum b_k z^k$$ where $$b_0 = 1, b_1 = -1$$ and $$b_k = 0$$ for $$k \ge 2$$, their Cauchy product will have the expansion: $$\sum c_k z^k$$ where the $$c_k$$ are given by $$c_k = \sum_{i+j=k} a_i*b_j$$.
Plugging in our values for the $$b_j$$ and evaluating the sum, we get exactly: $$c_0 = a_0 b_0 = 1$$ and $$c_k = a_0*b_1 + a_1*b_0 = 1-1=0$$ for $$k \geq 1$$ because $$b_j$$ is zero for $$j \geq 2$$
Thus, we conclude that the cauchy product is actually the series: $$f*g (z) = 1 + 0z + 0z^2 + ...$$ which in turn obviously converges for all values of $$z$$.