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?
On
The problem in the attempt is that we cannot use in general the ratio test because we are not sure that the terms are different from $0$.
Since the radius of convergence of $\sum_n b_nx^n$ is $2$, the series converges at $1$ hence $b_n\to 0$. This implies that the series $\sum_n |a_nx^n|$ is convergent if $|x|<1$.
Considering $a_n$ and $b_n$ such that $a_{2n}=0$, $a_{2n+1}=1$ and $b_{2n+1}=0$, $\sum_n b_{2n}x^n$ has a radius of convergence of 2., we can see that 1., 2. and 3. do not necessarily hold.
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 $x < 1$ 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 $\infty$.