Problem: If $\sum a_n z^n$ and $\sum b_n z^n$ have radii of convergence $R_1$ and $R_2$, show that the radius of convergence of $\sum a_n b_n z^n$ is at least $R_1 R_2$.

Is the following proof correct?


  1. We have by Hadamard's formula that

    $$ \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n}} =R_1 $$


    $$ \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|b_n \right|^{1/n}} = R_2 $$

  2. Consider now that

    $$ \underset{n \rightarrow \infty}{\limsup} \left|a_n b_n \right|^{1/n} \le \underset{n \rightarrow \infty}{\limsup} \left( \left|a_n \right| \left| b_n \right|\right)^{1/n} = \underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n} \cdot \underset{n \rightarrow \infty}{\limsup} \left|b_n \right|^{1/n} $$

  3. So that therefore

    $$ \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n b_n \right|^{1/n}} \ge \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left( \left|a_n \right| \left| b_n \right|\right)^{1/n} } = \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n}} \cdot \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|b_n \right|^{1/n}} = R_1 R_2 $$

  4. The last inequality then yields -- via Hadamard's formula -- that the radius of convergence for the power series $\sum a_n b_n z^n$ is at least $R_1R_2$, as desired.

Is this correct?


Not quite. In step 2., you have $$\limsup_{n\to\infty} \lvert a_nb_n\rvert^{1/n} = \limsup_{n\to\infty} \bigl(\lvert a_n\rvert\,\lvert b_n\rvert\bigr)^{1/n} \leqslant \limsup_{n\to\infty} \lvert a_n\rvert^{1/n}\cdot \limsup_{n\to\infty} \lvert b_n\rvert^{1/n}$$

instead of what you wrote, and correspondingly in step 3. you also need to swap the $=$ and $\geq$. After that, it will be correct.


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