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?
On
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 $.
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 $a_n$.