Let $R_1,R_2$ and $R$ be the radii of $\sum_{n=0}^\infty a_n z^n$, $\sum_{n=0}^\infty b_n z^n$ and $\sum_{n=0}^\infty a_nb_n z^n$ respectively. Prove or disptrove: $R_1R_2 \leq R$

$$\frac{1}{R_1} = \limsup_{n\to \infty}|(a_n)^{1/n}|$$

$$\frac{1}{R_2} = \limsup_{n\to \infty}|(b_n)^{1/n}|$$

and $$\frac{1}{R} = \limsup_{n\to \infty}|(a_n b_n)^{1/n}|=\lim_{n\to \infty} |(a_n)^{1/n}(b_n)^{1/n}|\leq \limsup_{n\to \infty}|(a_n)^{1/n}|\limsup_{n\to \infty} |(b_n)^{1/n}|=\frac{1}{R_1 R_2}$$

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    $\begingroup$ Your first two equalities for $R_1$ and $R_2$ are false. For example, you could have $a_n=n$ for all even $n$ and $a_n=1/n$ for all odd $n$... If you want formulas for the radii of convergece the best you have is the Cauchy-Hadamard formula, which involves a $\limsup$, and that is where you'll get the inequality. $\endgroup$ – Mariano Suárez-Álvarez Feb 5 '15 at 20:47

Let $\epsilon > 0$ be given. Then there exists $n > N$ such that $$ \lvert a_n\rvert^{1/n} < 1/R_1 + \epsilon,\qquad \lvert b_n\rvert^{1/n} < 1/R_2 + \epsilon $$ since $\limsup_{n\to\infty}\lvert a_n\rvert^{1/n} = 1/R_1$ so $\lvert a_n\rvert^{1/n} < 1/R_1 + \epsilon$ and similarly for $b_n$. Multiplying we obtain $$ \lvert a_nb_n\rvert^{1/n} < \frac{1}{R_1R_2} + \epsilon(1/R_1 + 1/R_2) + \epsilon^2 $$ Then $$ \frac{1}{R}\leq \frac{1}{R_1R_2}\Rightarrow R_1R_2\leq R $$


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