# $\sum_{n=1}^\infty a_n$ converges and $\sum_{n=1}^\infty|a_n|$ diverges. Then Radius of convergence?

Suppose $$\{a_n\}$$ be a sequence of real numbers such that $$\sum_{n=1}^\infty a_n$$ converges and $$\sum_{n=1}^\infty|a_n|$$ diverges. Then the radius of convergence of the power series $$\sum_{n=1}^\infty a_nx^n$$ .

Given answer: Radius of Convergence($$R)=1$$. My attempt:- There exists a natural number $$N$$ such that $$\forall n\ge N \implies |a_n|<1$$ (because $$\sum_{n=1}^\infty a_n$$ converges $$\implies \lim_{n\to \infty}a_n=0$$) So, $$\lim_{n\to \infty}|a_n|^{1/n}\le1\implies R\ge1$$)

Similarly, There exists a natural number $$M$$ such that $$\forall n\ge M \implies \sum_{k=1}^n|a_k|>1$$ ($$\because$$ $$\sum_{n=1}^\infty |a_n|$$ diverges ). How to complete the proof?

• $\sum_{n=1}^\infty a_n$ is NOT a power series. It makes no sense to talk about its radius of convergence. My guess is that you're asked for the radius of convergence of $\sum_{n=1}^\infty a_n x^n$ – jjagmath May 10 at 2:47
• Yes. Sorry for the Typo – Unknown x May 10 at 3:28

Some basic facts about power series: if $$R$$ is the radius of convergence of $$\sum a_n x^{n}$$ then the series converges absolutely for $$|x| and diverges for $$|x| >R$$. The first fact implies that $$R \leq 1$$ and the second fact implies that $$R\geq 1$$.
• Sir, Which factor implies $R\leq 1$.? I am not able to prove it. – Unknown x May 10 at 14:09
• If R >1 then the series will converge absolutely when $x=1$. – Kavi Rama Murthy May 10 at 14:30
• Sir, please see my attempt. I got $R\ge 1$. I don't get your point. How do I prove $R\leq 1$. Can you please help me? – Unknown x May 14 at 1:02
• @Unknownx If $R>1$ then $\sum a_n x^{n}$ converges absolutely for $|x| <R$, in particular for $x=1$. This is a basic fact about power series. Since it is given that $|sum a_n$ is not absolutely convergent it follows that $R$ cannot be greater than $1$. – Kavi Rama Murthy May 14 at 5:27