# computation of upper limit and lower limit

the infimum is when $n=1$, infimum is $-1$ the supremum is when $n=2$, supremum $1+\frac{1}{2}=\frac{3}{2}$

I need help to understand this part:

over the set $n \geq m$, the infimum is $\frac{-1}{2k+1}$ where $2m-1=2(2k+1)-1=4k-1$

or $2m+1=4k-1$

when $m \rightarrow ∞, k \rightarrow ∞$, so the infimum is tending to $0$

So the limit inferior is $0$

My problem is I do not know how to put it in notation of limit

over the set $n \geq m$, the supremum is $1+\frac {1}{2k}$ where $2m=4k$ or $2m+2=4k$

when when $m \rightarrow ∞, k \rightarrow ∞$, so the supremum decreases to $1$

So the limit superior is $1$ thanks for your help

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Can you typeset please? –  user17762 May 28 '12 at 17:15
It's really hard to understand your question. –  srijan May 28 '12 at 18:00
It seems the question (unrelated to limits) is to know how to write $-1/(2k+1)$ (first case) and $1+1/(2k)$ (second case) as functions of $m$ if $m=2k$ or $m=2k-1$. Is that so? –  Did May 28 '12 at 22:13
By definition, given any sequence $\{a_m\}_{m\in\mathbb{N}}$ of reals, $$\liminf_{m\to\infty}a_m:=\lim_{m\to\infty}\left(\inf_{n\geq m}a_n\right),$$ and $\limsup_{m\to\infty}a_m$ is similarly defined.
In this particular context (if I'm interpreting your informatiin and question correctly), given your sequence--which I will label $\{a_m\}_{m\in\mathbb{N}}$--we have $$\limsup_{m\to\infty}a_m=\lim_{m\to\infty}\left(1+\cfrac{1}{2\lceil m/2\rceil}\right)=1,$$ and the limit inferior can be similarly found.