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Given a series $\sum\limits_{n=0}^{\infty}a_nx^n$ with radius of convergence $R$, I need to find the radius of convergence of the series $\sum\limits_{n=0}^{\infty}a_n^3x^n$.

I would like to use d'Alembert theorem and say: $R'=\lim _{n \to \infty}\left|\frac{a_n^3}{a_{n+1}^3}\right|=R^3$. What's wrong with that? What is the correct answer?

Thank you!!

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The problem is that the limit $\frac{a_n^3}{a_{n+1}^3}$ may not exist, even if for $n$ large enough we have $a_n\neq 0$ (for example with $a_n=\sin n$. Maybe you can use the definition of the radius of converges, and show that if $|x|>R^3$ the series $\sum_{n=0}^{+\infty}a_n^3x^3$ diverges, and converges if $|x|< R^3$. –  Davide Giraudo Feb 6 '12 at 19:08
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Even simpler: use the fact that $a_nx^n\to0$ if $|x|\lt R$ and $a_nx^n\not\to0$ if $|x|\gt R$ (and that this characterizes $R$). –  Did Feb 6 '12 at 19:13
    
Why does $a_nx^n \to 0$ when $|x|>R$ ? and I don't understand where does it lead me to. –  Jozef Feb 6 '12 at 20:05
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When $|x|\gt R$, $a_nx^n$ DOES NOT converge to zero (as written in my last comment). You can use these as a characterization of $R$, that is, if you find $R'$ such that $a_n^3x^n$ converges to $0$ if $|x|\lt R'$ and $a_n^3x^n$ does not converge to $0$ if $|x|\gt R'$, then this $R'$ is your radius of convergence. –  Did Feb 6 '12 at 20:26
    
I'm sorry, I don't understand your answer. –  Jozef Feb 6 '12 at 20:32
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2 Answers

up vote 4 down vote accepted

Write $R(a)$ for the radius of convergence of the series $\sum\limits_na_nx^n$. Introduce $b_n=(a_n)^{\color{green}{\alpha}}$ for a given positive integer $\color{green}{\alpha}$ and write $R(b)$ for the radius of convergence of the series $\sum\limits_nb_nx^n$.

Lemma: Assume there exists a number $\varrho$ such that: (1) if $|x|\lt\varrho$, then $a_nx^n$ converges to $0$, and (2) if $|x|\gt\varrho$, then $a_nx^n$ does not converge to $0$. Then $R(a)=\varrho$.

Application:

  • Assume that $|x|\lt R(a)^\color{green}{\alpha}$. Then $|x|^{1/\color{green}{\alpha}}\lt R(a)$, hence $a_n|x|^{n/\color{green}{\alpha}}$ converges to $0$. Thus $b_nx^n$ converges to $0$ since $|b_nx^n|$ is the $\color{green}{\alpha}$th power of the modulus of $a_n|x|^{n/\color{green}{\alpha}}$.
  • Assume that $|x|\gt R(a)^\color{green}{\alpha}$. Then $|x|^{1/\color{green}{\alpha}}\gt R(a)$, hence $a_n|x|^{n/\color{green}{\alpha}}$ does not converge to $0$. Thus $b_nx^n$ does not converge to $0$ since $|b_nx^n|$ is the $\color{green}{\alpha}$th power of the modulus of $a_n|x|^{n/\color{green}{\alpha}}$.

Thus: $$\color{red}{R(b)=R(a)^\color{green}{\alpha}}.$$

Proof of the lemma: Assume that (1) and (2) hold.

  • Let $x$ such that $|x|\lt\varrho$. There exists $\sigma$ such that $|x|\lt\sigma\lt\varrho$. By (1), $a_n\sigma^n$ converges to $0$, in particular $|a_n\sigma^n|\leqslant C$ for every $n$. Thus $|a_nx^n|\leqslant Cu^n$ with $u=|x|/\sigma\lt1$, hence $\sum\limits_na_nx^n$ converges. Since this holds for every $|x|\lt\varrho$, $R(a)\geqslant\varrho$.
  • Let $x$ such that $|x|\gt\varrho$. By (2), $a_nx^n$ does not converge to $0$, hence $\sum\limits_na_nx^n$ diverges, which means that $R(a)\leqslant|x|$. Since this holds for every $|x|\gt\varrho$, $R(a)\leqslant\varrho$. QED.
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Thank you Didier. –  Jozef Feb 7 '12 at 10:25
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If you have the formula $R = \liminf_{n \rightarrow \infty} |a_n|^{-{1 \over n}}$, then the new radius of convergence is $\liminf_{n \rightarrow \infty} |a_n^3|^{-{1 \over n}} = \liminf_{n \rightarrow \infty} (|a_n|^{-{1 \over n}})^3 = (\liminf_{n \rightarrow \infty} |a_n|^{-{1 \over n}})^3 = R^3$.

But to use the above you should show that the liminf of the cubes of a sequence of nonnegative real numbers is the cube of the liminf.

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Thank you Zarrax. –  Jozef Feb 7 '12 at 10:26
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