Cauchy Product

• February 14th 2009, 08:15 PM
manjohn12
Cauchy Product
Suppose $\sum a_{n}z^{n}$ and $\sum b_{n}z^{n}$ have radii of convergence $R_1$ and $R_2$ respectively. Show that the Cauchy Product $\sum c_{n}z^{n}$ converges for $|z| < \min(R_1, R_2)$.

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

Is this correct?
• February 15th 2009, 03:55 AM
HallsofIvy
Quote:

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

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