Suppose that

  1. $a_n \geq 0$;
  2. $A_n=a_1+a_2+\cdots+a_n$;
  3. $\lim\limits_{n \to \infty}A_n=+\infty$;
  4. $\lim\limits_{n \to \infty}\dfrac{a_n}{A_n}=0.$

Show that the convergence radius of $\sum\limits_{n=1}^{\infty}a_nx^n$ equals $1$.


Let the convergence radius of $\sum\limits_{n=1}^{\infty}a_nx^n$,$\sum\limits_{n=1}^{\infty}A_nx^n$ be $r,R$ respectively. If there exists a $x>0$ such that $\sum\limits_{n=1}^{\infty}A_nx^n$ converges, according to that $0 \leq a_nx^n \leq A_nx^n$, by the comparison test for positive series, we have $\sum\limits_{n=1}^{\infty}a_nx^n$ is convergent. Hence $r \geq R.$ Moreover, from $\lim\limits_{n \to \infty}\dfrac{a_n}{A_n}=0$, we may obtain $\lim\limits_{n \to \infty}\dfrac{A_{n+1}}{A_n}=1.$ Thus, $R=1$. Notice that $A_n$ is the partial sum of $\sum\limits_{n=1}^{\infty}a_n$. Since $\lim\limits_{n \to \infty}A_n=+\infty$, we have $\sum\limits_{n=1}^{\infty}a_n$ is divergent. Thus, $r \leq 1.$ As a result, $r=1.$

Am I right?

  • $\begingroup$ $A_n x^n$ in the third line should read $a_n x^n$. $\endgroup$ – Florian R May 30 '18 at 8:10
  • 1
    $\begingroup$ You're proof seems correct to me. The convergence is interval is thus $[-1,1)$. $\endgroup$ – Václav Mordvinov May 30 '18 at 8:22
  • $\begingroup$ @FlorianR yes,I correct that. $\endgroup$ – mengdie1982 May 30 '18 at 8:22

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