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Let $\sum c_n z^n$ be a power series. ($c_n,z_n \in \mathbb{C}$)

Let $\alpha = \lim \sup {|c_n|}^{1/n}$ Then this series is convergent if $|z|<1/{\alpha}$.

Let $\beta = \lim \sup |c_{n+1}/{c_n}|$ Here, series is convergent if $|z|<1/{\beta}$.

Since $\alpha≦\beta$, $|z|<1/{\alpha}$ gives more choice of $|z|$ which makes the series convergent.

It seems like the second one is easier to apply to problems. Is any constraint that makes $\alpha=\beta$?

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    @Ilya Already changed it when right after I saw my comment, stupid programming habit!2012-08-24

2 Answers 2

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The second is indeed easier to apply, however, you should be careful, since that second limit is only defined if for every $n$, $c_n\neq 0$.

HINT: If $ \lim |c_{n+1}/c_n|$ exists then $\alpha = \beta$.

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According to Rudin, Principles of Mathematical Analysis, Thm 3.37, for any sequence $c_n$ of positive numbers, $\liminf_{n\rightarrow \infty} \frac{c_{n+1}}{c_n} \le \liminf_{n\rightarrow \infty} (c_n)^{1/n}$ while $\limsup_{n\rightarrow \infty} (c_n)^{1/n} \le \limsup_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}$ As mentioned by others care has to be taken when some $c_k$ are zero.

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    This inequality was already mentioned at MSE a few times, sometimes with a proof, too; e.g. [here](http://math.stackexchange.com/questions/69386/inequality-involving-limsup-and-liminf) and [here](http://math.stackexchange.com/questions/28476/finding-the-limit-of-frac-n-sqrtnn/28487#28487). (And in questions linked to those ones.)2012-08-24