More specifically, if $\sum a_nx^n$ is a power series then the Cauchy–Hadamard theorem has me computing $\lim_{n\to\infty}\sup\left(\vert a_n\vert^{1/n}\right)$. Why is the supremum operator there? If the limit exists then it exists and if it blows up to infinity then it blows up to infinity, what does the supremum add to it? I'm only use the supremum being used in the context of sets.
What does $\lim_{n\to\infty}\sup(f(n))$ mean?
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calculus
real-analysis
analysis
limits
power-series
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4It should be $\limsup_{n\to\infty}|a_n|^{\frac{1}{n}}$. Do you know what the limsup of a sequence is? – 2017-01-11
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1If not, see here: https://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior In practice, the limit often exists anyway – 2017-01-11
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1The limit might not exist even though the limsup does exist. – 2017-01-11
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0@Brian Borchers could you give me an example of such a case please? – 2017-01-11
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1If $a_{2k}=1$ and $a_{2k+1}=0$ then $\lim |a_n|^{1/n}$ does not exist, but $\limsup |a_n|^{1/n}=1$. @David – 2017-01-11
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0@Thomas Andrews got it, thank you. – 2017-01-11
1 Answers
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Limit superior is defined as follows:
$$\limsup_{n\to\infty}f(n)=\lim_{n\to\infty}\sup_{m\ge n}f(m)$$
where $\sup_{m\ge n}f(m)$ means to take the maximum value $f(m)$ takes for $m\ge n$.
The importance is that it ignores other sub-sequences, which might make the limit not exist. For example, if every odd term was $0$, it wouldn't affect convergence, but it would affect the limit.
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0Also importantly, $\limsup$ always exists (though it might be one of $\pm\infty$.) – 2017-01-11