It does seem like a very simple question, but I don't see how to do it properly.

How can we show that the limit of


as $j\rightarrow\infty$does not exist, while $\limsup$ and $\liminf$ exist and are equal to $1$ and $\frac{1}{3}$?

I can see it intuitively but don't see how to prove it. Thank you very much for any help.

  • $\begingroup$ If the limit exists, then $\limsup$ and $\liminf$ must be equal. $\endgroup$ – Jonas Granholm Apr 13 '14 at 12:20
  • $\begingroup$ I don't know how to prove that $\limsup$ and $\liminf$ are what they are, so that is where I'm stuck $\endgroup$ – user135450 Apr 13 '14 at 12:21
  • $\begingroup$ To be able to see it, you can interpret these liminf/limsup as : for $N$ as big as I want, I can find $n \geq N$ such that $u_n$ is arbitrary close to liminf (or limsup) (show it from the definition). So if $u_n$ would have a limit $l$, for $n$ big enough, you must have $u_n$ arbitrary close to $l$,limsup and liminf at the same time. Since the three can't be equal ($liminf \neq limsup$), there is a contradiction (it's the same idea as in the proof of the unicity of the limit) $\endgroup$ – yago Apr 13 '14 at 12:26
  • $\begingroup$ Should the $k$ in $3^{\lfloor\frac{-\log(k)}{\log(2)}\rfloor}j^{\frac{\log(3)}{\log(2)}}$ be a $j$? $\endgroup$ – David H Apr 13 '14 at 12:33
  • $\begingroup$ Yes indeed, sorry typo :) $\endgroup$ – user135450 Apr 13 '14 at 13:09

Every term of the sequence is in $(\frac13,1]$ since $x-1\lt\lfloor x\rfloor\leqslant x$ for every $x$. If $j=2^n$, the result is $1$. If $j=2^n+1$, the result is $\frac13\,\left(1+\frac1{2^n}\right)$. Hence the limsup is $1$ and the liminf is $\frac13$.


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