For conditionally convergent series $\sum a_n,\exists (m_n)_{n\in\mathbb{N}}$ with $(n-1)k<m_n\leq n k,$ s.t. $\sum_{n\in\mathbb{N}} a_{m_n}=\alpha.$

Question

Properties of conditionally convergent series $\ \displaystyle\sum a_n\ $:

1. The sum of the positive terms is $+\infty;\ $ the sum of the negative terms is $-\infty.$
2. $\displaystyle\lim_{\substack{ { n\to\infty } \\ a_n\geq 0 }} a_n = 0^+\ ;\ $ $\displaystyle\lim_{\substack{ { n\to\infty } \\ a_n < 0 }} a_n = 0^-. $

Proposition:

Let $\ \displaystyle\sum a_n\ $ be a conditionally convergent series, let $\ k\in\mathbb{N}_{\geq 2},\ $ and let $\ \alpha\in\mathbb{R}.\ $ Then $\ \exists\ (m_n)_{n\in\mathbb{N}} \subset \mathbb{N}\ $ with $\ (n-1)k < m_n \leq nk\ \forall\ n\in\mathbb{N},\ $ such that $\ \displaystyle\sum_{n\in\mathbb{N}} a_{m_n} = \alpha.$

I have thought about it quite a bit, but I'm not sure if the proposition is true or false. Any ideas for a proof or a counter-example?

Answer 1

There is a counterexample.

Let $a_n$ be a sequence from alternating harmonic series, but repeating numbers:

$$ 1, 1, -\frac 12, -\frac 12, \frac 13, \frac13, -\frac14, -\frac14, \cdots $$

Let $k=2$ and $\alpha=100$. In each interval $((n-1)k,nk]$, there is only one number among the repeated numbers is chosen for $a_{m_n}$. That is, $a_{m_n}$ takes the usual alternating harmonic sequence $$ 1,-\frac 12, \frac 13, -\frac14, \cdots $$ Thus, we can only have $$ \sum_{n=1}^{\infty} a_{m_n} = \log 2 \neq \alpha. $$