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My textbook defines: \begin{equation} \limsup (a_n) = \min\{M ∈ R |\ \ ∃n_0 \ \ ∀n > n_0, a_n ≤ M\}. \end{equation} And it gives an example: Let $a_n$= 1+ $\frac{1}{n}$. Then $\limsup (a_n)$ = 1.

This confuses me - under no circumstances 1 is bigger or equal to an element of the sequence. It's alawys a bit smaller than $a_n$. What's my mistake?

blz
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    Take any number $1 + \alpha > 1$ for some $\alpha > 0$. Can you find $n_0$ s. t. $1 + 1/n < 1 + \alpha$ for all $n > n_0$? Now, what is the infimum of the set ${ 1 + \alpha \mid \alpha > 0 }$? – Singhal Dec 07 '14 at 08:57
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    Your mistake is imagining that the textbook definition says that the $\limsup$ is greater than or equal to some term of the sequence. It says nothing of the kind. It says that $\limsup(a_n)$ is the greatest lower bound of some set. Why don't you start by figuring out what set that is in the given example; just which real numbers $M$ have the stated property? – bof Dec 07 '14 at 08:58
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    @godingly $M\neq 1$ but the infimum of $M$ is one. $\inf\neq\min$. – Suzu Hirose Dec 07 '14 at 09:09
  • Well, the textbook says you can write \begin{equation} \limsup (a_n) = \min {M ∈ R |\ \ ∃n_0 \ \ ∀n > n_0, a_n ≤ M}. \end{equation}. Then in that case, M $\neq$ 1 under any circumstance, and we're in the same problem. – blz Dec 07 '14 at 09:10
  • @Suzu Hirose However they define limsup as minimum of that set. So if 1 is the limsup, it must be in that set, but that's impossible. – blz Dec 07 '14 at 09:11
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    @godingly Stop messing around with editing comments and read the responses. Three people have answered your question now. – Suzu Hirose Dec 07 '14 at 09:12
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    I must admit that to silently change inf into min in the question after users explained the matter, is one of the most unconstructive attitudes one can imagine. – Did Dec 07 '14 at 09:28
  • Does your textbook say "inf" or "min"? Why won't you identify the book? – bof Dec 07 '14 at 21:35

2 Answers2

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Imagine that $\lim \sup (a_n) = y > 1$ then there exists some $n\geq 1/(y-1)$ such that $1+1/n<y$. This contradicts that $y$ is lower than any $M\in\Bbb R$ which is a supremum of $a_n$.

Suzu Hirose
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If the definition is as in your new edit, with min instead of inf, the you are right: the number doesn't exist for the sequence you give.

The common definition uses inf, which guarantees that the limsup always exists. In such case the limsup of your sequence is indeed $1$, as it is the infimum of the numbers $1+1/n $ where each of them is an upper bound for the tail of the sequence starting at $n $.

Martin Argerami
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