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$\sum_{n=0}^{\infty} a_nx^n$ has radius of convergence $R_{1}$ and $\sum_{n=0}^{\infty} b_nx^n$ has radius of convergence $R_{2}$. Show that $\sum_{n=0}^{\infty} a_nb_nx^n$ has radius of convergence $\ge$ $R_{1}R_{2}$.

First of all, can I say that $\sum_{n=0}^{\infty} a_nx^n$ has radius of convergence $R_{1}$ $\implies$ $\frac{1}{\limsup_{n\to \infty} \lvert a_n\rvert^{\frac{1}{n}}}=R_{1}$ ? I know the implication works in the other direction but I'm not fully sure about this direction.

Assuming the forward direction works, we have $${\limsup_{n\to \infty} \lvert a_n\rvert^{\frac{1}{n}}}={\frac{1}{R_{1}}}$$ and $${\limsup_{n\to \infty} \lvert b_n\rvert^{\frac{1}{n}}}={\frac{1}{R_{2}}}$$

Then $\lvert a_n\rvert^{\frac{1}{n}}\lvert b_n\rvert^{\frac{1}{n}}=\lvert a_nb_n\rvert^{\frac{1}{n}}$ $\implies$ $\frac{1}{\limsup_{n\to \infty} \lvert a_nb_n\rvert^{\frac{1}{n}}}=R_{1}R_{2}$ $\implies$ $\sum_{n=0}^{\infty} a_nb_nx^n$ has radius of convergence exactly equal to $R_{1}R_{2}$ (but we want to show r.o.c. $\ge$ $R_{1}R_{2}$).

unless for some reason that I don't see $\lvert a_n\rvert^{\frac{1}{n}}\lvert b_n\rvert^{\frac{1}{n}}\ge\lvert a_nb_n\rvert^{\frac{1}{n}}$ I have no idea how to show r.o.c. $\ge$ $R_{1}R_{2}$ instead of r.o.c. = $R_{1}R_{2}$.

I also can't come with any examples where $\sum_{n=0}^{\infty} a_nx^n$ has r.o.c. $R_{1}$ and $\sum_{n=0}^{\infty} b_nx^n$ has r.o.c. $R_{2}$ but $\sum_{n=0}^{\infty} a_nb_nx^n$ has r.o.c. strictly > $R_{1}R_{2}$


GEdgar: i thought that, but then i thought: ${\limsup_{n\to \infty} \lvert a_n\rvert^{\frac{1}{n}}}$ = ${\inf_{n>0}\sup_{m>n} \lvert a_n \rvert^{\frac{1}{n}}}$ and ${\limsup_{n\to \infty} \lvert b_n\rvert^{\frac{1}{n}}}$ = ${\inf_{n>0}\sup_{m>n} \lvert b_n\rvert^{\frac{1}{n}}}$

and also generally, sup(AB)=sup(A)sup(B) and inf(AB)=inf(A)inf(B) when elements of A and B are nonnnegative. so ${\limsup_{n\to \infty} \lvert a_n\rvert^{\frac{1}{n}}\lvert b_n\rvert^{\frac{1}{n}}}$ = ${\inf_{n>0}\sup_{m>n} \lvert a_n\rvert^{\frac{1}{n}}\lvert b_n\rvert^{\frac{1}{n}}}$ = ${\inf_{n>0}(\sup_{m>n} \lvert a_n\rvert^{\frac{1}{n}} \sup_{m>n} \lvert b_n\rvert^{\frac{1}{n}})}$ = $({\inf_{n>0}\sup_{m>n} \lvert a_n \rvert^{\frac{1}{n}}})$ $({\inf_{n>0}\sup_{m>n} \lvert b_n\rvert^{\frac{1}{n}}})$ = ${\limsup_{n\to \infty} \lvert a_n\rvert^{\frac{1}{n}}}$ ${\limsup_{n\to \infty} \lvert b_n\rvert^{\frac{1}{n}}}$

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  • $\begingroup$ I think you can show that it converges for $x=(R_1-\epsilon)(R_2-\epsilon)$ with the Dirichlet test. $\endgroup$ – Gregory Grant Nov 11 '15 at 18:24
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    $\begingroup$ If by $AB$ you mean $\{ab \colon a \in A, \, b \in B\}$ then note that this is not what you have when you take $a_nb_n$ where you only pair up specific elements, not all pairs. $\endgroup$ – quid Nov 11 '15 at 19:01
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Product of the limsups may not be limsup of the product. But there is an inequality...

For example if $x_n$ alternates $0,1$ but $y_n$ alternates $1,0$ then $x_ny_n=0$ while both limsups are $1$.

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  • $\begingroup$ my comment was too long so i edited it in at the end of the post $\endgroup$ – Baroque Spiderman Nov 11 '15 at 18:56
  • $\begingroup$ got it. thanks! $\endgroup$ – Baroque Spiderman Nov 11 '15 at 19:06
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$x + x^3 + x^5 + \cdots $ and $x^2 + x^4 + \cdots$ both have radius of convergence $1.$ In this case $a_nb_n=0$ for all $n,$ so the ROC for $\sum a_nb_n$ is $\infty.$

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I think you can show that it converges for $x=(R_1-\epsilon)(R_2-\epsilon)$ with the Dirichlet test.

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