Attempt:
Write $\dfrac{c_n}{c_{n+1}} = \dfrac{ a_0 b_n + a_1 b_{n-1} + … + a_nb_0 }{a_0b_{n+1} + a_1 b_n + … + a_{n+1} b_0} $
I can see this approach may seem a little laborious. I was trying to compute $\lim c_n/c_{n+1}$ and show that this is $R$ or at least $\leq R$. Any hints?
As for the secont part,notice that
$$ f(z) g(z) = (a_0 + a_1 z + a_2 z^2 + …. )(b_0 + b_1 z + b_2 z^2 + … ) = a_0b_0 + (a_0 b_1 + b_0 a_1 + a_0b_0 ) z + … $$
So, it is clear that eventually $c_n$ will have the same representation that is given. Now, how do I make this argument formal? I mean it seems obvious…
Best Answer
By Mertens Theorem if the series $\sum_{n=0}^{\infty}x_n\rightarrow x$ and $\sum_{n=0}^{\infty}\rightarrow y$ and the latter series converges absolutely, then their Cauchy product $$\sum_{n=0}^{\infty}z_n=\sum_{n=0}^{\infty}\sum_{k=0}^{n}x_ky_{n-k}\rightarrow xy$$. Applying this to the power series' $f(z)$ and $g(z)$ you can conclude that the radius of convergence is at least $R$.
For a proof of Mertens theorem see https://en.m.wikipedia.org/wiki/Cauchy_product#Convergence_and_Mertens.27_theorem
Let $f(z)$ be the Taylor series expansion of $\frac{1}{\sqrt{1-x}}$ about $0$ and $g(z)$ be the Taylor series expansion of $\sqrt{1-x}$ about $0$. Both have radius of convergence $1$ but their product is the constant power series $1$ and hence has radius of convergence $\infty$.