A restricted version of Riemann series theorem: rearrangements with alternating signs

Question

If $(a_{n})$ is a conditionally convergent series in real field, then for any real number $\alpha$, there exists a rearrangement $(a_{k_{n}})$ of $(a_{n})$ such that for all even $n$, $a_{k_{n}} \geq 0$, for all odd $n$, $a_{k_{n}} \leq 0$, and $\sum a_{k_{n}} = \alpha$.

This problem boils down to the following problem: can every conditional real series be rearranged to an alternative convergent series. If this is solved, then apply the extensions of Riemann's theorem by Sierpiński, the original problem is done.

I appreciate any suggestion about this problem. Thanks in advance.

Answer 1

Here you prescribe in addition the sequence of signs of the rearranged series in the Riemann-Dini theorem to be alternating, but note that any non-stationary binary sequence of signs does as well. More precisely:

Let $(a_k)_{k\in\mathbb N} $ be an infinitesimal sequence of non-zero real numbers such that $\sum_{k\ge0}a^+_k=\sum_{k\ge0}a^-_k=+\infty$.

Let $\epsilon\in \{-1,1\}^\mathbb{N}$ be a non-eventually constant sequence.

Let $\alpha\in\mathbb R \cup\{ \pm\infty\}$.

Then there exists a permutation $\sigma$ of $\mathbb N$ such that $$\sum_{k=0}^\infty a_{\sigma(k)}=\alpha$$ $$\text{sgn}\,a_{\sigma(k)}=\epsilon_k.$$

To this end: extract a subset $S\subset\mathbb N$ such that $\sum_{k\in S}a_k$ is absolutely summable and $a_k$ are positive resp. negative for infinitely many $k\in S$ (therefore for infinitely many $k\in \mathbb N\setminus S$ as well, because of the assumption $\sum_{k\ge0}a^+_k=\sum_{k\ge0}a^-_k=+\infty$).

Then do the Riemann-Dini bijection $\tau:\mathbb N\to \mathbb N \setminus S$ relatively to the sequence $(a_k)_{k\in \mathbb N\setminus S }$ and the number $\alpha-\sum_{k\in S}a_k$, namely $$\sum_{k=0}^\infty a_{\tau(k)}=\alpha-\sum_{k\in S}a_k.$$

Finally, insert the coefficients $\{a_k\}_{k\in S}$ in some order into the series $\sum_{k=0}^\infty a_{\tau(k)}$, so as to get a rearrangement with the prescribed final sequence of signs $(\dagger)$. Since $\sum_{k\in S}a_k$ is absolutely summable, the order of the insertion does not affect the convergence and the value of the sum, which is $\alpha$ as wanted.

$(\dagger)$ This can easily be done, for the reason that any non eventually constant binary sequence contains any other non eventually constant binary sequence as a subsequence, in such a way that the complement is also non eventually constant.

Answer 2

I believe the answer is yes. Let $a_{i_k}$ and $a_{j_k}$ be the even and odd terms respectively, ordered in increasing order of magnitude.

We will define four sequences; $b_n$ (indexed by $\mathbb Z_+$) which will be the desired rearranged sequence, a $-1, 1$ valued sequence $s_n$ (indexed by $\mathbb N$), which will be a “state/control” variable, and two sequences $P_n, M_n$ (indexed by $\mathbb N$) of natural numbers.

To this end, consider the following algorithm:

Start algorithm.

1. Set $P_0 = M_0 = 0$.

2. If $|a_{i_0}| \geq |a_{j_0}|$ , set $s_0 = 1$, else set $s_0 = -1$.

3. Assume $s_0, \dots, s_{n-1}; b_1, \dots, b_{n-1}, P_{n-1}, P_{n-1}$ have been defined,

4. Set $P_{n}$ (respectively $M_{n})$ to be the largest index $k$ such that $a_{i_m}$ (respectively $a_{j_m}$) has already been used for all $m < k$.

5. Do the following:

While $s_{n-1} = 1$, and $n$ is odd,

• If $\overset{n-1}{\underset {i=1} {\sum}} b_r >\alpha - a_{i_{P_{n}}}$, set $b_n = a_{j_{M_n}}$ and set $s_n = 1$; else set $b_n = a_{j_l}$ for any $l \geq M_n$ with $|a_{j_l}| \leq \frac{1}{2}|a_{i_{P_n}}|$ and set $s_n = -1$.

While $s_{n-1} = 1$, and $n$ is even,

• If $\overset{n-1}{\underset {i=1} {\sum}} b_r >\alpha - a_{i_{P_{n}}}$, set $b_n = a_{i_l}$ for any $l$ with $|a_{i_l}|\leq \frac{1}{2} |a_{M_n}|$ and set $s_n = 1$; else set $b_n = a_{i_{P_n}}$ and set $s_n = -1$.

While $s_{n-1}= -1$, and $n$ is even,

• If $\overset{n-1}{\underset {i=1} {\sum}} b_r < \alpha - a_{i_{M_{n}}}$, set $b_n = a_{i_{P_n}}$ and set $s_n = -1$; else set $b_n = a_{i_l}$ for any $l \geq P_n$ with $|a_{i_l}| \leq \frac{1}{2}|a_{j_{M_n}}|$ and set $s_n = 1$.

While $s_{n-1} = -1$, and $n$ is odd,

• If $\overset{n-1}{\underset {i=1} {\sum}} b_r < \alpha - a_{j_{M_{n}}}$, set $b_n = a_{j_l}$ for any $l$ with $|a_{j_l}|\leq \frac{1}{2} |a_{P_n}|$ and set $s_n = -1$; else set $b_n = a_{i_{M_n}}$ and set $s_n = 1$.

End algorithm.

I claim this algorithm exhausts all terms. Indeed it suffices to show that $b_n = a_{i_{P_n}}$ and $b_n = a_{j_{P_n}}$ infinitely often. We prove only the former, the proof for the latter being identical via symmetry.

Thus assume for contradiction that $b_k = a_{i_{P_k}}$ for only finitely many $k$, and let $K$ be the largest of these. Then the terms $b_n$ starting from $n = K+1$ are $a_{i_{M_n}}, a_{i_{l_n}}, a_{i_{M_{n+1}}}, a_{i_{l_{n+1}}}, \dots$, and further it is always the case that $\overset{n-1}{\underset {i=1} {\sum}} b_r >\alpha - a_{i_{P_{n}}}$ for even $n > K$.

But since $|a_{i_{l_n}}| \leq \frac{1}{2}|a_{i_{M_n}}|$, and the sequence $\sum a_{j_k}$ diverges (to negative infinity), we must eventually have $\overset{n-1}{\underset {i=1} {\sum}} b_r < \alpha - a_{i_{P_{n}}}$, contradiction.

Finally, to see that $\sum b_n$ converges to $\alpha$, it suffices to note that by construction, once $s_n$ has switched values at least once, then from that point on $\overset{n}{\underset {i=1} {\sum}} b_r$ is no more than $2(|a_{P_n}| + |a_{M_n}|)$ away from $\alpha$, which goes to 0 as $n \to \infty$.