# Difference between limsup and sup

I'm having some difficulty visualising the difference between the limit supremum and supremum (and for limit infimum/infimum) for bounded sequences. Would it be possible for some to provide a brief explanation and maybe some examples?

• Sup you can calculate to sets, not to sequences – Martín Vacas Vignolo Apr 9 '16 at 1:10
• @vvnitram can you elaborate a bit? – Inazuma Apr 9 '16 at 1:11

## 1 Answer

Let $$(x_n)$$ be a bounded sequence. The supremum is the the least upper bound of the sequence as a set. For the limit-supremum, or $$\limsup$$, is the limit of the sequence $$(a_n)$$, where $$a_n=\sup_{k\geq n}x_k$$. In plainer language, the limit supremum measures how the supremum of the sequence behaves as we start removing terms from the sequence, starting from the beginning. The nice thing about $$\limsup$$ is that it always exists, even if the sequence doesn't converge at all.

Sometimes the $$\limsup$$ is equal to the $$\sup$$. Take for example the sequence $$x_n=1-1/n$$, where $$\sup\{x_1,x_2,x_3,...\}=1$$. If we take the $$\sup$$ of the subsequence $$\{x_n,x_{n+1},x_{n+2},...\}$$, we still get $$1$$ for each $$n$$, so $$\limsup\{x_n\}=1$$.

Other times, $$\limsup$$ is less than $$\sup$$. If $$x_n=1/n$$, $$\sup\{x_1,x_2,x_3,...\}=x_1=1$$. If we remove the first term, $$\sup\{x_2,x_3,x_4,...\}=x_2=1/2$$. Similarly, the more terms we remove, the lower the $$\sup$$ is. As we remove more and more terms, the $$\limsup\{x_n\}=0$$. The $$\limsup$$ is never greater than $$\sup$$. See if you can prove this to yourself.