# Radius of convergence of a power series with $a_n$ convergent

Let $\{a_n:n \geq 1\}$ be a sequence of real numbers such that $\sum_{n=1}^{\infty} a_n$ is convergent and $\sum_{n=1}^{\infty} |a_n|$ is divergent. Let $R$ be the radius of convergence of the power series $\sum_{n=1}^{\infty} a_n x^n$. Then what is $R$ ?

I dont know how to start, if the series is given we could use Radius of convergence formula, but how to solve this question?Pls help

• Well it can't be bigger than $1$, and it can't be less than $1$, so it's $1$. – Gregory Grant May 24 '15 at 7:06
• Yes..You are right.The answer is 1. How did you find, please explain – Sam Christopher May 24 '15 at 7:07
• It converges for $x=1$, so do you see why it can't be less than $1$. – Gregory Grant May 24 '15 at 7:08
• Think about the series $\sum (-1)^n\frac{1}{n}$ – Empty May 24 '15 at 7:11

It converges for $x=1$ so the radius cannot be less than one. Now we know it converges absolutely for $x$ in the interior of the interval, and the condition shows it converges conditionally for $x=1$, so the radius cannot be greater than one (if it were, then $x=1$ would be in the interior and it would have to converge absolutely). So it must be exactly one.