Radius of $\sum a_n b_n x^n$ via radii of $\sum a_n x^n$ and $\sum b_n x^n$ Series $\sum a_n x^n$ and $\sum b_n x^n$ have radii of convergence of
  1 and 2, respectively. Then radius of convergence R of $\sum a_n b_n x^n$ is
  
  
  2
  1
  $\geq 1$
  $ \leq 2$
  


My attempt was to apply ratio test: 

$$\lim_{n \rightarrow \infty} \frac{a_{n} b_{n}}{a_{n+1} b_{n+1}} = \lim_{n \rightarrow \infty} \frac{a_{n}}{a_{n+1}} \cdot \lim_{n \rightarrow \infty} \frac{ b_{n}}{ b_{n+1}} = 1 \cdot 2 = 2$$ i.e. option (1).

But the answer is (3). Why?
You can't apply the limit test because you don't know the limits $|\frac{a_n}{a_{n+1}}|$ or $|\frac{b_n}{b_{n+1}}|$. If these limits exist, you can use them to find a RoC, but knowing the radius doesn't imply that the limits exist.*

To show the answer is $3$, recall that a power series converges absolutely (and uniformly) inside it's radius of convergence. 

Take any $x &lt; 1$ and let $S(x) = \sum |a_nx^n|$. For any $n$ $|a_nb_nx^n|\leq S(x)|b_n|$ and $\sum S(x)|b_n|$ converges (since $1$ is inside the radius of convergence for $\sum b_nx^n$), so by the comparison test, $\sum a_nb_nx^n$ converges. Thus the radius of convergence is at least $1$.

Technically, this only proves that $3)$ is a possible answer. You would have to show that $1)$, $2)$ and $4)$ don't hold, by finding examples of power series that exclude these possibilities. It would in fact be enough to find a series that has RoC larger than $2$, since this would simultaneously exclude everything other than $3)$ (and if you assume that one of the answers is correct, this would actually be enough to prove that $3$ is the correct answer.)**

*Take for example, any convergent power series and replace the even coefficients with $0$. Clearly this gives a convergent power series, with RoC at least as large as the original, but the ratio test fails since the limit isn't defined.

**For an example of this, take $a_n = 1$ if $n$ is even, $0$ if $n$ is odd and $b_n = \frac{1}{2}^n$ if $n$ is odd, $0$ if $n$ is even. These have radii of convergence $1$ and $2$ respectively, as required, but $a_nb_n$ is identically $0$, so has raidus of convergence $\infty$.
we are given that $\limsup\sqrt[n]{|a_n|}=1$ and $\limsup\sqrt[n]{|b_n|}=\frac12$. Show that this implies $\limsup\sqrt[n]{|a_nb_n|}\le \frac12$.