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$\{s_n\}$ is defined by $$s_1 = 0; s_{2m}=\frac{s_{2m-1}}{2}; s_{2m+1}= {1\over 2} + s_{2m}$$

The following is what I tried to do.

The sequence is $$\{0,0,\frac{1}{2},\frac{1}{4},\frac{3}{4},\frac{3}{8},\frac{7}{8},\frac{7}{16},\cdots \}$$

So the even terms $\{E_i\} = 1 - 2^{-i}$ and the odd terms $\{O_k\} = \frac{1}{2} - 2^{-k}$ and each of them has a limit of $1$ and $\frac{1}{2}$, respectively.

So, the upper limit is $1$ and the lower limit is $1\over 2$, am I right ?

Does this also mean that $\{s_n\}$ has no limits ?

Is my denotation $$\lim_{n \to \infty} \sup(s_n)=1 ,\lim_{n \to \infty} \inf(s_n)={1 \over 2} $$ correct ?

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    $\begingroup$ Looks right to me! $\endgroup$ – Peter Košinár May 31 '13 at 23:26
  • $\begingroup$ Answers to your questions in order: yes, yes and yes. (but it is "notation", not "denotation") $\endgroup$ – DonAntonio May 31 '13 at 23:39
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Shouldn't it be $E_i = \frac{1}{2} - 2^{-i}$ and $O_i = 1 - 2^{1-i}$? That way $E_i = 0, \frac{1}{4}, \frac{3}{8}...$ and $E_i = 0, \frac{1}{2}, \frac{3}{4}...$, which seems to be what you want.

Your conclusion looks fine, but you might want to derive the even and odd terms more rigorously. For example, the even terms $E_i$ are defined recursively by $E_{i+1} = s_{2i+2} = \frac{s_{2i+1}}{2} = \frac{E_1 + \frac{1}{2}}{2}$, and $\frac{1}{2} - 2^{-i}$ also satisfies this recursion relation. $E_1 = 0$, and $\frac{1}{2} - 2^{-1} = 0$, hence they have the same first term. By induction the two sequences are the same.

If we partition a sequence into a finite number of subsequences then the upper and lower limit of the sequence are equal to the maximum upper limit and minimum lower limit of the subsequences; in this case you're partitioning into even and odd terms.

$\{s_n\}$ has a limit iff the upper and lower limits are the same (this is proved in most analysis books), so in this case $\{s_n\}$ has no limits.

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  • $\begingroup$ Thank you. Unfortunately my book (Rudin) says "it's obvious that $\{s_n\}$ has a limit iff the upper and lower limits are the same" so it never stuck to me. $\endgroup$ – hyg17 Jun 1 '13 at 0:50

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