# How to explain the following definition of $\lim\sup s_n$ intuitively?

How to explain the following definition of $\lim\sup s_n$ intuitively?

let $\lim\sup s_n$=$l$

The definition is ($\forall t<l,\forall N,\exists n>N, s_n>t$) and ($\forall t>l,\exists N,\forall n>N, s_n<t$)

For the first one, if we have a increasing sequence from very negative to $0$, how is it still true. i.e. $\{-1, -0.5, ... , -0.00001,...\}$, here $\lim\sup s_n=0$. let $t=-1,$ so $t<0$, but not all terms are greater than $t$.

For the second one, I am also confused. I think It would be great if someone could show some valid examples.

Also when we prove the such limit exists, is it necessary to show both characteristics?

• The first says: if $t < l$, then infinitely many members of the sequence are greater than $t$. The second says: if $t > l$, then for $n$ sufficiently large, $s_n < t$. – Daniel Schepler May 23 '17 at 23:47
• Who writes symbolic definitions like these unless one is doing this in a course on formal logic? Definitions are meant to be understood properly and not to be stored in cryptic fashion. It is a simply an ineffective definition if it is not understood easily and clearly. A much better approach to these concepts is given in this answer math.stackexchange.com/a/1893725/72031 – Paramanand Singh May 24 '17 at 4:27

Informally speaking the $\limsup s_n$ is the limit of the sequence defined from the supremum of the tails of the sequence $(s_n)$.

Here the intuitive key is the concept of tail of a sequence. A tail, as it descriptive name want to define, is a sequence resulting of discard the first $m$ elements of the sequence (where $m$ is finite, consequently a tail ever have infinite elements).

We will represent the sequence visualized as

$$\overbrace{s_0,s_1,\underbrace{s_2,\ldots,s_k,\ldots}_{\text{a tail of }(s_n)}}^{\text{the sequence }(s_n)}$$

Hence we can define the sequence of tails from the original sequence $(s_n)$ as

$$t_0=s_0,s_1,\ldots$$

$$t_1=s_1,s_2,\ldots$$

$$t_k=s_k,s_{k+1},\ldots$$

Then $\limsup s_n=\lim(\sup t_n)$.

Your first criterion means that for all $t<l$, there exists a subsequence $s_{n_k}$ of $s_n$ such that all members of the subsequence are greater than $t$.

Your second criterion means that for every $t>l$ there exists a point in the sequence $s_n$ from which all sequence members are smaller than $t$.