Radius of convergence powers series s.t seriex $ \sum |a_n| $diverges [UGC-CSIR-NET 2014 December]

Let $\{a_n:n\ge 1\}$ be sequence of real numbers such that the series
  $\sum a_n$ converges and the series $\sum |a_n|$ diverges. Let $R$ be
  the radius of convergence of the power series $\sum a_n x^n$; then
  which of the following is true.
  
  
*
  
*$0\lt R \lt 1$
  
*$1\lt R \lt \infty$
  
*$R=1$
  
*$R=\infty$

In this question, I am a little confused regarding the role of the series $\sum a_n$ in its power series $\sum a_n x^n$ in earlier questions that I have faced, I find the radius of convergence of absolute version of coefficient of power series, where absolute version converges for those 'x' values origional power series also converges. Is that right?
But as we see here absolute version diverges, so what does it imply we can't use ratio, root test etc to find radius of convergence here?
 A: Lemma. 
Let the power series $\sum a_nx^n$ converge at some point $x=x_0$. 
Then it converges absolutely for all $x$ satisfying $|x|<|x_0|$. That is,
$$
\sum |a_n||x|^n < \infty ,
$$
whenever $|x|<|x_0|$.
Proof. Suppose that the power series $\sum a_nx^n$ converges at some $x=x_0$. Then necessarily $|a_nx_0^n|\to0$ as $n\to\infty$. In particular, there is some $M$ such that $|a_n| |x_0|^n\leq M$ for all $n$, or $|a_n|\leq M |x_0|^{-n}$ for all $n$.
Now let $x$ be such that $|x|<|x_0|$. Then 
$$
|a_nx^n|\leq |a_n||x|^n \leq M \big(\frac{|x|}{|x_0|}\big)^n = M\rho^n ,
$$
where $\rho=\frac{|x|}{|x_0|}<1$, which shows that $\sum a_nx^n$ converges absolutely. QED.
Returning back to the problem, the series $\sum a_n$ converges means that the power series $\sum a_nx^n$ converges at $x=1$. Hence by the lemma it converges for all $x$ with $|x|<1$, implying that the convergence radius $R$ cannot be less than $1$. i.e., $R\geq1$. 
Now suppose $R>1$, that is, the power series converges at some $x=x_0$ with $|x_0|>1$. Then by the lemma, the power series must converge absolutely at $x=1$, which means that $\sum|a_n|<\infty$. But we have $\sum|a_n|=\infty$, so $R$ must be equal to $1$.
A: Here are some properties of the radius of convergence $R$:


*

*If $|z|<R$ then the series $\sum |a_n z^n|$ is convergent ( that is, $\sum a_n z^n$ is absolutely convergent)

*If $|z| >R$ then the series $\sum a_n z^n$ is not convergent. 
Equivalently
1' If $\sum_n |a_n z^n|$ is divergent then $|z|\ge R$
2' If $\sum_n a_n z^n $ convergent then $|z|\le R$
Now we have the conditions on the LHS in 1', 2' for $z=1$. Therefore we have also the RHS conditions, that is $1\ge R$ and $1\le R$, and thus $R=1$.
