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A sequence $\{b_n\}_{n=1}^∞$ is a reordered sequence of $\{a_n\}_{n=1}^∞$.
A series $\sum_{n=1}^∞ a_n$ converges not absolutely but conditionally.
[ ] refers to floor function.
$S_n=\sum_{k=1}^n b_k$.
$d_n=S_n-[S_n]$
Then, I want to prove that a set $\{d_n\ ; n∈$N$\}$ is dense on the interval $[0,1)$.

I thought that Riemann series theorem is useful, but I don’t know how to use it in this proof.

And I thought that in order to prove the density, I should first prove that there exists $d_n$ such that $ \frac{k}{2^m} ≦ d_n <\frac{k+1}{2^m}$.
$m$ is an arbitrary natural number and k is an arbitrary natural number ($0≦k≦2^m-1$).

However, I can’t even prove this.

Please help me.

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    $\begingroup$ Some elements in your question require clarification. Absolute convergence is defined for series, not for sequences. Hence can you revisit your sentence the séquence $\{a_n\}$ converges absolutely? Also, even if you replace sequences with series, the result isn’t true in general. $\endgroup$ – mathcounterexamples.net Jul 1 '18 at 13:09
  • $\begingroup$ It is not clear what is meant by "a set $\{ S_n - [S_n] \}$ is dense". That set, as it written now, has only one element, thus it is not dense. Is it indexed by $n$ (i.e. it denotes a sequence), or is it indexed by all reorderings of $\{ a_n \}$ (or something else)? I think that solving this ambiguity is the first step to get a solution. $\endgroup$ – Crostul Jul 1 '18 at 13:13
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    $\begingroup$ If the sequence $b_n$ is fixed, it is possible that it is fixed to a permutation of $a_n$ for which $\sum_{n\geq0} b_n$ is also convergent with sum $S$. In that case, $S_n-[S_n]$ is a convergent sequence tending to either $S-[S]$, if $S$ is not an integer, or having two accumulation points $0,1$ if $S$ is an integer. It can't be dense. Now, if we are allowed to change $b_n$, then by Riemann's theorem we can choose $S$ to be any number in $[0,1)$. Then $S_n-[S_n]$ tends to $S-[S]=S$. $\endgroup$ – user566930 Jul 1 '18 at 13:32
  • $\begingroup$ @mathcounterexamples.net Thank you. I editted it. I thought this is true, but if it's false, I would like $\endgroup$ – Gymnast Jul 1 '18 at 13:41
  • $\begingroup$ you to give a counterexample. $\endgroup$ – Gymnast Jul 1 '18 at 13:42
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We prove the following, which in particular, covers the claim of the original question:

Claim: Let $b_n$ be a sequence of reals, such that $b_n \to 0$ but $S_n = \sum_{k=1}^n b_k \to +\infty$. Then the sequence $\{S_n - [S_n] \}_{n=1}^\infty$ is dense in $(0,1)$.

Proof: The idea is simple: since the sequence $S_n$ diverges to $+\infty$ with decaying steps $b_n$, it will eventually enter intervals of any length between integers (since it's speed of diverging to infinity becomes smaller and smaller as $n$ grows; sketching this on a line should make things very clear). We now formally exploit this idea.

Fix $\varepsilon>0$ small, and let $N\in \mathbb{N}$ be so that $|b_n| \leq \varepsilon$ for all $n>N$. Assume also that $[S_{n_1}] = k$ where $n_1>N$ is fixed. This means $$ k\leq S_{n_1} < k + 1. $$ Recall that $S_n \to +\infty$, hence for some $n_2> n_1$ we must have $S_{n_2}>k+2$. But since $|b_n|<\varepsilon$ for all $n>n_1$ it follows that any interval in $(k+1, k+2)$ of length at least $2\varepsilon$ must contains at least 1 member of the sequence $S_n$. This is because $|S_n - S_{n-1}| = |b_n| < \varepsilon$, so $S_n$ cannot jump over an interval of length $2\varepsilon$ in $1$ step. Getting back to fractional parts, the latter shows that any interval of length at least $2\varepsilon$ in $(0,1)$ contains an element of the sequence $\{S_n - [S_n]\}$. Since $\varepsilon>0$ was arbitrary the claim follows.

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