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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
up vote 2 down vote accepted

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 information 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.

Is that what you were looking for? If not, let me know. It is usually helpful to those who would answer you to give as much of the problem's context as possible. As srijan noted, it's difficult to tell what you are asking.

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