Prove that $\{S_n - [S_n]\}$ is dense on the interval $[0,1)$ when $S_n$ is $\sum_{k=1}^n b_k$. 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.
 A: 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.
