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I don't really understand the concept of Lim Sup and why it is any different from a standard limit. I am working on a question about radius of convergence and I know that $r = \frac{1}{\limsup\limits_{n\to\infty} A_n^\frac{1}{n}}$ but don't have any intuitive idea of what this means. Could someone put it into words that I can understand please? Thanks and sorry for my inability to format things correctly on here!

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  • $\begingroup$ limsup$_{n \to \infty} a_n$ is the limit of the sequence $s_n := sup \{a_k : k \geq n\}$. If the limit $\lim_{n \to \infty} a_n$ exists, then the limsup is the same as the limit. It's useful in situations where the limit doesn't exist. For example, if $a_n = (-1)^n$, it doesn't make sense to talk about the limit of the sequence $(a_n)$, but the limit of the associated sequence $s_n$ is 1 (since $s_n = 1$ for all $n$). $\endgroup$ – ah11950 Apr 11 '14 at 14:49
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Difference of $\limsup$ and $\lim$

If you have a series that is not convergent like this one:

$$ s_n := \left\{\begin{matrix} \frac{1}{n} & n \text{ is even}\\ 5+\frac{1}{n^2} & n \text{ is not even} \end{matrix}\right.$$

Than your Limes just does not exists, in a lot of cases it is still possible to deduce information from such series. By just taking the "biggest limes" or biggest limit point.

So whats the $\limsup$ here basically:

$$ \begin{align*} \limsup_{n\to\infty} s_n &= 5 \\ \liminf_{n\to\infty} s_n&= 0\end{align*}$$

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  • $\begingroup$ so if the sequence does not converge you can still find out how high or low it goes? is that the idea here? $\endgroup$ – Mmm Apr 11 '14 at 14:58
  • $\begingroup$ yep, even thought the limsup might also not converge in some cases. $\endgroup$ – mjb4 Apr 11 '14 at 14:59
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Consider the sequence

$$ 0, 1.1, 0, 1.01, 0, 1.001, 0, 1.0001, 0, 1.00001, 0, \ldots $$

This series is clearly divergent, but there is still information content in its limiting behavior: e.g. it has two limit points, $0$ and $1$. The lim sup of this sequence is $1$, and the lim inf is 0.

I pick this example, because it demonstrates we need something more sophisticated than simply the supremum of the sequence, which is $1.1$. The lim sup can be interpreted as the limiting behavior of "the supremum of the tail".

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Let $\{a_n\}$ be any sequence. Look at the supremum of the sequence starting at the first term, and then at the second, and so on: \begin{align*} c_1 &= \sup \{a_1, a_2, a_3,\ldots\} \\ c_2 &= \sup \{a_2, a_3, a_4, \ldots\} \\ c_3 &= \sup \{a_3, a_4, a_5, \ldots\} \end{align*} and so on. If $c_1 = \infty$ then in fact $c_n = \infty$ for all $n$ so that $c_n \to \infty$. On the other hand, if $c_1 < \infty$ we have $c_1 \ge c_2 \ge c_3 \ge \cdots$. Since any nonincreasing sequence converges (possibly to $-\infty$) the sequence $\{c_n\}$ converges. Thus $\lim_{n \to \infty} c_n$ makes sense regardless of whether or not $\{a_n\}$ converges. The $\limsup$ of $\{a_n\}$ is just this limit.

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