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?

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    $\begingroup$ $\sum_{n=1}^\infty a_n$ 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 $\sum_{n=1}^\infty a_n x^n$ $\endgroup$ – jjagmath May 10 '19 at 2:47
  • $\begingroup$ Yes. Sorry for the Typo $\endgroup$ – user464147 May 10 '19 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 $|x| <R$ and diverges for $|x| >R$. The first fact implies that $R \leq 1$ and the second fact implies that $R\geq 1$.

  • $\begingroup$ Sir, Which factor implies $R\leq 1$.? I am not able to prove it. $\endgroup$ – user464147 May 10 '19 at 14:09
  • $\begingroup$ If R >1 then the series will converge absolutely when $x=1$. $\endgroup$ – Kavi Rama Murthy May 10 '19 at 14:30
  • $\begingroup$ 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? $\endgroup$ – user464147 May 14 '19 at 1:02
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    $\begingroup$ @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 $1$. $\endgroup$ – Kavi Rama Murthy May 14 '19 at 5:27
  • $\begingroup$ Thank you very much. $\endgroup$ – user464147 May 14 '19 at 5:52

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