Cauchy Product

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?

Quote:

---

Originally Posted by 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?

---

Yes, it is. Notice, however, that this does NOT say that the radius of convergence is min(R1,R2). It is possible that the Cauchy product of two power series converges outside the radius of convergence of one of them.