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Find all subsequential limits of: $$ x_n=\left\{1,{1\over2},{2\over2},{3\over2},{1\over4},{4\over4},\dots,{9\over 4},\dots,{1\over2^n},{2\over2^n},\dots,{3^n\over2^n},\dots\right\} $$

I've been recently solving a similar problem. After that I've been considerting the following sequence: $$ y_n = \left\{1, {1\over 2}, {2\over2},{3\over2},{1\over 4},{2\over 4},{3\over4},{4\over4},{5\over 4},\dots,{1\over 2^n},\dots,\frac{2^n - 1}{2^n},\dots \right\} $$

Using a similar technique from the linked answer one may introduce a sequence: $$ r_n = \frac{\lfloor r2^n\rfloor}{2^n} $$ And then: $$ r2^n - 1 < \lfloor r2^n\rfloor \le r2^n $$ Dividing both parts by $2^n$: $$ r - {1\over 2^n} < r_n \le r $$ Which by squeeze theorem gives that: $$ \lim_{n\to\infty}\left(r - {1\over 2^n}\right) < \lim_{n\to\infty}r_n \le \lim_{n\to\infty} r = r $$ Now if we take $r\in [0, 1]$ then $r_n$ should be somewhere in the original $y_n$. Thus $r$ is a subsequentilal limit. Hence the whole set of reals $[0, 1]$ is the set subsequential limits.

If i understand this correctly $y_n$ is nothing but splitting the range $[0, 1]$ into infinitely many nested intervals. Which eventually end up in a single point belonging to each interval. And we can choose any point within that interval appearing to a subsequential limit.

However I'm not sure how to apply this reasoning to $x_n$. Intuitively it seems like $[0, +\infty]$ is the set of subsequential limits, but I'm having a hard time proving this.

I'm kindly asking to help me out with that.

roman
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  • If you could prove that your set is dense in $\mathbb{R_+}+{\infty}$, then your claim follows. Also, following the same approach why not consider $\dfrac{\lfloor ra^n\rfloor}{a^n}$ where $a = \frac 32 ?$ – dezdichado Dec 13 '18 at 18:24
  • For a given $r$ and integer $N$ then there is an integer $M$ such that $|r - \frac{M}{2^N}| < \frac{1}{2^N}$. If $N$ is large enough then $M < 3^N$. Prove this (start from $0$ and go towards $r$ with steps of size $1/2^N$). Use this to construct a subsequence that converges to $r$. – Winther Dec 13 '18 at 18:30

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