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Question: Suppose that ($a_n$) is a sequence such that L = lim sup ($a_n$) is a real number. Then we know that, for any number $M$ $>$ $L$, there are only finitely many integers $n$ for which ($a_n$) $>$ $M$. Show by means of an example that it is possible to have ($a_n$) $>$ $L$ for infinitely many n.

I don't really understand what the question means, any further explanation would be really appreciated. There is a Proposition in my textbook that I think may be helpful for this question, but I am not sure.

  • Proposition 2.20

(a) If $L$ = $lim$ $sup$ ($a_n$) then

  • for each $ε$ $>$ 0 the inequality ($a_n$) $>$ $L$ + $ε$ holds for only finitely many n and
  • For each $ε$ $>$ 0 the inequality ($a_n$) $>$ $L$ - $ε$ holds for infinitely many n

Any help would be really appreciated

Thanks a lot

Joy Yin
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  • HINT: If $a_n=\frac1n$, what is $\limsup_na_n$? – Brian M. Scott Jan 24 '16 at 20:57
  • Would the $lim$ $sup$ ($a_n$) be 0? – Joy Yin Jan 24 '16 at 20:59
  • It would indeed. – Brian M. Scott Jan 24 '16 at 21:00
  • Sorry, what is this question asking? I don't quite understand what example I need to show. – Joy Yin Jan 24 '16 at 21:02
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    It’s simply asking for a sequence that has infinitely many terms that are larger than the lim sup. It’s asking for this because if you take any number bigger than the lim sup, only finitely many terms of the sequence can be larger than that number. In the example that I suggested, if $b>0$, only finitely many of the terms $\frac1n$ are greater than $b$, but all of them are greater than the lim sup $0$. – Brian M. Scott Jan 24 '16 at 21:04

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