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Let consider the series $$\sum_{n\in\mathbb Z}a_nz^n.$$ We denote $R$ the radius of $\sum_{n=0}^\infty a_nz^n$ and $r$ the radius of $\sum_{n=-\infty }^{-1}a_nz^n$, i.e.e the series converge absolutely if $r<|z|<R$. The thing I don't understand is why $$\frac{1}{r}=\liminf_{n\to \infty }\left|\frac{a_{-n-1}}{a_{-n}}\right|.$$ Indeed, $$\sum_{n=-\infty }^{-1}a_nz^n=\sum_{n=1}^\infty a_{-n}z^{-n},$$ and thus, by d'Alembert, it converge if $$\limsup_{n\to \infty }\left|\frac{a_{-n-1}}{a_{-n}}\right|\frac{1}{|z|}<1\implies |z|>\limsup_{n\to \infty }\left|\frac{a_{-n-1}}{a_{-n}}\right|:=r.$$

Therefore $$r=\limsup_{n\to \infty }\left|\frac{a_{-n-1}}{a_{-n}}\right|.$$

What's wrong here ? Why this $\liminf$ ? I can't get it.

vidyarthi
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user330587
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    Both are incorrect. If $$\lim_{n\to\infty} \biggl\lvert \frac{a_{-n-1}}{a_{-n}}\biggr\rvert$$ exists, then $r$ is that limit. But if the limit doesn't exist, $r$ need not be connected to the $\liminf$ or $\limsup$. Consider $a_{-n} = \frac{1}{n^2}$ for odd $n$ and $a_{-n} = \frac{1}{n^3}$ for even $n$. Then $r = 1$, but the $\liminf$ is $0$, and the $\limsup$ is $+\infty$. – Daniel Fischer Jun 16 '17 at 11:40
  • @DanielFischer: Thanks. So is there a general formula ? And do you agree that $R=\limsup_{n\to \infty }\left|\frac{a_n}{a_{n+1}}\right|$ for $\sum_{n\geq 1}a_nz^n$ ? – user330587 Jun 16 '17 at 13:44
  • That suffers from the same problem. The general formula is the Cauchy-Hadamard formula. – Daniel Fischer Jun 16 '17 at 14:00

1 Answers1

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This is wrong:

$$\frac{1}{r}=\liminf_{n\to \infty }\left|\frac{a_{-n-1}}{a_{-n}}\right|$$

A better formula is

$$\frac{1}{r}=\liminf_{n\to \infty }\left|\frac{a_{-n}}{a_{-n-1}}\right|$$

But as Daniel Fischer points out in a comment, this is not generally correct. It is correct if $|a_{-n}|$ is monotonic.

TonyK
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