Rearranging a sequence of bounded variation

Question

Suppose $(c_j)$ is a sequence of complex numbers indexed by $j \in \mathbb{Z}$, which is of bounded variation in that $$ \sum_{j \in \mathbb{Z}} \lvert c_j - c_{j - 1} \rvert < \infty. $$ Then, does it follow that the infinite sum $$ \lvert c_1 - c_0 \rvert + \lvert c_{-1} - c_1 \rvert + \lvert c_{2} - c_{-1} \rvert + \lvert c_{-2} - c_{2} \rvert + \lvert c_{3} - c_{-2} \rvert + \ldots \text{ and so on} $$ is finite as well?

My attempts to bound the terms in the latter sequence by the boundedness of the sum of the former one have lead to too many factors of "$n$"! Thanks so much for the help.

Answer 1

How about

$$c_j = \begin{cases} 0 & \text{for } j \leqslant 0 \\ 1 & \text{for } j > 0 \end{cases}$$

Answer 2

No. Consider $d_n = 2^{-n}$, $c_0 = 0$, $c_n = d_1 + \dots + d_n$ and $c_{-n} = -c_n$ for all $n \in \Bbb{N}$.

$$\sum_{j \in \mathbb{Z}} \lvert c_j - c_{j - 1} \rvert = \sum_{j \in \mathbb{N}} d_j + \sum_{j \in \mathbb{N}} d_{j-1} < \infty,$$

but each term in $$\lvert c_1 - c_0 \rvert + \lvert c_{-1} - c_1 \rvert + \lvert c_{2} - c_{-1} \rvert + \lvert c_{-2} - c_{2} \rvert + \lvert c_{3} - c_{-2} \rvert + \cdots \text{ and so on}$$

is greater than $|c_1-c_0| = 1$, so the above sum diverges.