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Let $\{c_n\}$ be a sequence of complex numbers, and $R:=\sup\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}$, which means that R is the radius of convergence of the series $\sum\limits_{n=0}^\infty c_n(z-z_0)^n$ about $z_0$. Show that $R=\lim\inf\limits_{n\to\infty}|c_n|^{-1/n}$.

Unfortunately, I have no idea how to prove this. In fact, the very concept of the limit inferior is very new to me, and this concept is not immediately easy to completely grasp. Moreover, why is $\sup\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}\ne \max_{r\in\mathbb{R}}\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}$, or is it?

I would appreciate some guidance with this problem. In this case I feel completely lost.

1 Answers 1

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1) $R$ is the radius of convergence:
Proof.
Take $x$ with $0 Take $x$ with $x>R$. Then, by the definition of $R$, $\{c_nx^n\}$ is not bounded, so $c_nx^n\to 0$ $(n\to \infty)$ does not hold and we see that the series does not converge.
Thus $R$ is the radius of convergence.

2) $R=\liminf\limits_{n\to\infty}|c_n|^{-1/n}$
EDIT 2
I misunderstood what the author of the book intends. He asks for a direct proof of $$ R:=\sup\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}=\liminf_{n\to\infty}\frac{1}{\sqrt[n]{|c_nx^n|}}. $$ Proof. Take an arbitrary $0\le r Theorem. If $r \le a_n\implies r \le \liminf_{ n\to \infty} a_n.$

Tendeng $r \to R$ we have \begin{align} R\le \liminf_{n\to\infty}\frac{1}{\sqrt[n]{|c_n|}}.\tag{1} \end{align}

Next suppose that $$\rho :=\liminf_{n\to\infty}\frac{1}{\sqrt[n]{|c_n|}}>R.$$ Then there exists $N$ such that $$ \frac{1}{\sqrt[n]{|c_n| }}>\frac{\rho +R}{2}$$ holds for all $n>N$. Then we have $$ 1>|c_n|\left(\frac{\rho +R}{2} \right)^n.$$

This is a contradiction. Because $|c_n|\left(\frac{\rho +R}{2} \right)^n$ is bounded and $\frac{\rho +R}{2}>R$. Thus we can conclude \begin{align} \liminf_{n\to\infty}\frac{1}{\sqrt[n]{|c_n|}}\le R\tag{2} \end{align} and $(1)$ $(2)$ implies $$ R=\liminf_{n\to\infty}\frac{1}{\sqrt[n]{|c_n|}}.$$

3) $\sup\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}\ne \max_{r\in\mathbb{R}}\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}$:
Example.
Take $c_n=n.$ Then $\{c_nr^n\}$ is bounded for $r$ with $r<1$ and hence
$$\sup\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}=1.$$ On the contrary, $\max_{r\in\mathbb{R}}\{r\ge 0: \{c_nr^n\} \text{ is bounded}\}$ does not exist. Indeed $$ \{r\ge 0: \{c_nr^n\} \text{ is bounded}\}=[0,1).$$

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    In one book, I read that the set $\{c_nr^n: 0\le r – 2017-02-17
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    Can you please clarify how come $\frac{1}{\lim\sup\limits_{n\to\infty}|c_n|^{1/n}}=\lim\inf\limits_{n\to\infty}|c_n|^{-1/n}$? – 2017-02-18
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    To first comment: What we can say definitively are \begin{align} &\text{I. } \sum_{n=1}^\infty c_n r^n \text{ converges} \implies \lim_{ n\to \infty} c_nr^n=0\implies \{c_nr^n\} \text{ is bounded}.\\ &\text{II. } \{c_nr^n\}\text{ is not bounded} \implies\lim_{ n\to \infty} c_nr^n\ne 0\implies \sum_{n=1}^\infty c_n r^n \text{ does not converge}. \end{align} The boundedness of $\{c_nr^n\}$ does not imply the convergence of $ \sum_{n=1}^\infty c_n r^n$. Even if $\{c_nr^n\}$ is bounded, sometimes $ \sum_{n=1}^\infty c_n r^n$ diverges and sometimes $ \sum_{n=1}^\infty c_n r^n$ converges. – 2017-02-18
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    To the second comment: In general $$\frac{1}{\lim\sup\limits_{n\to\infty}|c_n|^{1/n}}=\lim\inf\limits_{n\to\infty}|c_n|^{-1/n}$$ does not hold. I will write the proof of 2) later. – 2017-02-18
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    But $R$ is the radius of convergence of the infinite sum. So $\{c_nr^n:0\le r< R\}$ must necessarily be bounded I think. But the book says that the definition of $R$, that is the radius of convergence of the sum, does not necessarily imply $\{c_nr^n:0\le r< R\}$ is bounded. How so? – 2017-02-18
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    I've written a more comprehensive question about this theorem at http://math.stackexchange.com/questions/2149502/cauchy-hadamard-confusion – 2017-02-18
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    added a proof of 2) – 2017-02-18