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This is quoted from wikipedia,

The limit superior of $x_n$ is the smallest real number $b$ such that, for any positive real number $\epsilon$ , there exists a natural number $N$ such that $x_n < b + \epsilon$ for all n > N . In other words, any number larger than the limit superior is an eventual upper bound for the sequence. Only a finite number of elements of the sequence are greater than $b + \epsilon$ .

Now doesn't all the elements lie within $b$ i.e. $\forall x \in <x_n> , x \leq b $ ? Isn't the limit superior the upper bound of the sequence? If so, why should there be any elements greater than $b$ as said in the bolded statement of wikipedia above? How can there be elements, though finite, greater than $b$? Where am I mistaking? Please help as I am new in this topic. Thanks in advance.

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    $\begingroup$ Consider the sequence $88, 1/2, 3/4,7/8,15/16,\dots$. The limsup is $1$. $\endgroup$ May 18 '15 at 7:49
  • $\begingroup$ @André Nicolas: Hi, sir. Yes the element of the sequence like $15/16$ is greater than the limsup. So, basically what does limsup actually tell? I thought it as the upper bound. Was I wrong? $\endgroup$
    – user142971
    May 18 '15 at 7:59
  • $\begingroup$ Yes, you are wrong. It is the largest limit point. $\endgroup$ May 18 '15 at 11:40
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    $\begingroup$ There is only one lub (sup). The limsup is sometimes equal to the sup, but often different and smaller. $\endgroup$ May 18 '15 at 11:54
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    $\begingroup$ Yes, you have the idea, you can think of it as the sup of the accumulation points. $\endgroup$ May 18 '15 at 12:38
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The LIMIT superior is the key. Note, $\limsup s_n=\limsup s_{n+k}$ for all $k$. Be careful with your definition. If it was just the "sup" of the sequence, then you would be correct. But it is the LIM sup.

The definition of lim sup is $\limsup\limits_{n\to\infty} s_n=\lim\limits_{n\to\infty} \sup \{ s_k\colon k>n\} $. So if $\limsup s_n =L$, there can be finitely many points$p_1,\cdots,p_l$ greater than $L$, because there is some $n$ such that you only consider the supremum of the set AFTER that.

Edit, just as a tip for analysis. Analysis is the study of infinities. So as a general rule, finitely many things don't matter.

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  • $\begingroup$ Thanks sir for posting the answer. Can you tell me what is the difference between sup & limsup & least upper bound of a sequence? I'll be grateful to you:) $\endgroup$
    – user142971
    May 18 '15 at 8:04
  • $\begingroup$ Sir, can you tell what you meant by $\lim{\sup}s_n = \lim{\sup}s_{n+k}$? $\endgroup$
    – user142971
    May 18 '15 at 11:40
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    $\begingroup$ I just gave the definition of limsup. You already know what sup (which is a synonym for least upper bound) means. $\endgroup$
    – user223391
    May 18 '15 at 18:55
  • $\begingroup$ Yes sir, but if sup is synonymous to least upper bound, then limsup actually does implies there must be a convergent sequence of sup whose limit is limsup, right? So, is there actually a sequence of upper bounds? After reading this , I think there exits such sequence but they are actually accumulation points... $\endgroup$
    – user142971
    May 19 '15 at 4:29
  • $\begingroup$ ...But do you think this is right ie. the sup points are actually accumulation points? And the limit of the sequence of those accumulation points is the limsup? $\endgroup$
    – user142971
    May 19 '15 at 4:59
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The limit superior is the limit of the sequence derived from $(a_n)_{n\ge 0}$ by starting further and further; explicitly, it is the sequence: $$b_0=\sup_{n\ge 0}a_n,\enspace b_1=\sup_{n\ge 1}a_n,\enspace b_2=\sup_{n\ge 2}a_n,\dots$$ which is a decreasing sequence.

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