Rearrange conditionnaly convergent sequence so the partial sums form a dense subset of $\mathbb{R}$

Question

I was reading about conditional convergence vs absolute convergence. I understand from the wikipedia on Riemann series theorem how we construct a permutation such that the partial sum converges to any real or to infinity. I also read this post that sais that there is a rearrangement for which the partial sums are dense in $\mathbb{R}$. But how exactly do we construct the rearrangement?

Answer 1

First, a classical lemma: if a sequence $(u_k)$ has subsequential limits $a < b$ and $\lim_{k \to + \infty} (u_{k+1} - u_k) = 0$, then any points in $[a,b]$ is also a subsequential limit. Now, note that if we denote the series by $\sum_{k \geq 1} a_k$, then $a_k \to 0$ since the series is convergent, and therefore, it is easy to see that for any bijection $\varphi : \mathbb{N} \to \mathbb{N}$, we have $$ \lim_{k \to + \infty} a_{\varphi(k) = 0}. $$ Therefore, it suffices to construct a rearrangement whose partial sums have limsup $+ \infty$ and liminf $- \infty$.

To do this, take some upper bound $M$ for $(|a_n|)$ (possible since it tends to $0$ so is bounded). The fact that a series $\sum_{n \geq 1} a_n$ is conditionally convergent implies that $$ \sum_{i \geq 1} a_{n_i} = + \infty, \quad \sum_{i \geq 1} a_{m_i} = - \infty, $$ where $(n_i)$ (resp. $(m_i)$) are the indices corresponding to the positive (resp. strictly negative) terms. We can therefore find $i_1$ such that $$ \sum_{i = 1}^{i_1} a_{n_i} \in [1,1+M], $$ and we let $\varphi(i) = n_i$ for $i \in \{1, \dots, i_1\}$. Then, we can find $i_2$ such that $$ \sum_{i = 1}^{i_2} a_{m_i} \in [-2M-3, -M-3], $$ and we let $\varphi(i_1+i) = m_i$ for $i \in \{1, \dots, i_2\}$. In particular, we have $$ \sum_{i=1}^{i_1} a_{\varphi(i)} > 1, \quad \sum_{i=1}^{i_1+i_2} a_{\varphi(i)} < 1+M - M - 3 = -2. $$ Continuing, we obtain a permutation $\varphi$ and a sequence $(i_k)$ such that, if we let $$ S_k = \sum_{i=1}^k a_{\varphi(i)}, \quad j_k = i_1 + \cdots + i_k, $$ then for every $k \in \{1,2, \dots\}$, $$ S_{j_{2k-1}} > 2k-1, \quad S_{j_{2k}} < -2k. $$ In particular, $$ \limsup_{k \to + \infty} S_k = + \infty, \quad \liminf_{k \to + \infty} S_k = - \infty, $$ and we are done.