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**manjohn12** Suppose $\displaystyle \sum a_{n}z^{n} $ and $\displaystyle \sum b_{n}z^{n} $ have radii of convergence $\displaystyle R_1 $ and $\displaystyle R_2 $ respectively. Show that the Cauchy Product $\displaystyle \sum c_{n}z^{n} $ converges for $\displaystyle |z| < \min(R_1, R_2) $.

So $\displaystyle \sum c_nz^{n} = \sum_{k=0}^{n} a_{k}b_{n-k}z^{n}$. Assume $\displaystyle |z| < \min(R_1, R_2) $. Put $\displaystyle |z| = \min(R_1, R_2) - 2 \delta $ where $\displaystyle \delta < \min(R_1, R_2) $. Then $\displaystyle \overline{\lim}|c_n|^{1/n} = \frac{1}{\min(R_1,R_2)} $ for $\displaystyle n $ large enough. Thus $\displaystyle |c_{n}z^{k}| \leq \left(\frac{\min(R_1,R_2)-2 \delta}{\min(R_1,R_2)- \delta} \right)^{k} $. Thus $\displaystyle \sum c_{n}z^{n} $ converges.

Is this correct?