The Comparison Test shows that if is an absolutely convergent series, and is a bounded sequence, then converges absolutely. Indeed, where is such that for all .

With a bit more effort one can prove that this property of *preserving absolute convergence* is equivalent to being a bounded sequence. Indeed, if is unbounded, then for every there is such that . We can ensure since there are infinitely many candidates for . Define if for some , and otherwise. Then converges but diverges because its terms do not approach zero.

What if we drop “absolutely”? Let’s say that a sequence *preserves convergence of series* if for every convergent series , the series also converges. Being bounded doesn’t imply this property: for example, does not preserve convergence of the series .

**Theorem**. A sequence preserves convergence of series if and only if it **has bounded variation**, meaning converges.

For brevity, let’s say that is BV. Every bounded monotone sequence is BV because the sum telescopes. On the other hand, is not BV, and neither is . But is for . The following lemma describes the structure of BV sequences.

**Lemma 1.** A sequence is BV if and only if there are two increasing bounded sequences and such that for all .

**Proof.** If such exist, then by the triangle inequality and the latter sums telescope to which has a limit as since bounded monotone sequences converge.

Conversely, suppose is BV. Let , understanding that . By construction, the sequence is increasing and bounded. Also let ; as a difference of bounded sequences, this is bounded too. Finally,

which shows that is increasing.

To construct a suitable example where diverges, we need another lemma.

**Lemma 2.** If a series of nonnegative terms diverges, then there is a sequence such that the series still diverges.

**Proof.** Let (partial sums); then . The sequence tends to infinity, but slower than itself. Let , so that , and we are done: the partial sums of telescope to .

**Proof of the theorem, Sufficiency part**. Suppose is BV. Using Lemma 1, write . Since , it suffices to prove that and converge. Consider the first one; the proof for the other is the same. Let and write . Here converges as a constant multiple of . Also, converges by the Dirichlet test: the partial sums of are bounded, and decreases to zero.

**Proof of the theorem, Necessity part**. Suppose is not BV. The goal is to find a convergent series such that diverges. If is not bounded, then we can proceed as in the case of absolute convergence, considered above. So let’s assume is bounded.

Since diverges, by Lemma 2 there exists such that and diverges. Let be such that ; that is, differs from only by sign. In particular, . Summation by parts yields

As , the left hand side of does not have a limit since diverges. On the other hand, since while stays bounded. Therefore, does not exist.

Let . The series converges (by telescoping, since exists) but diverges, as shown above.

In terms of functional analysis the preservation of absolute convergence is essentially the statement that . Notably, the norm of , i.e., , is the number that controls by how much can exceed .

I don’t have a similar quantitative statement for the case of convergence. The BV space has a natural norm too, (interpreting as ), but it’s not obvious how to relate this norm to the values of the sums and .

Discrete BV has come up in a few Banach algebra papers from the 1960s/70s, I think, as it is an example of an l1 convolution algebra associated to a certain semigroup.

I’m not sure if this addresses your final remarks, but I think BV is the dual of c, the space of convergent sequences, provided that we define the pairing by mapping a convergent sequence to so that the paring is given by