# Multipliers preserving series convergence

The Comparison Test shows that if ${\sum a_n}$ is an absolutely convergent series, and ${\{b_n\}}$ is a bounded sequence, then ${\sum a_nb_n}$ converges absolutely. Indeed, ${|a_nb_n|\le M|a_n|}$ where ${M}$ is such that ${|b_n|\le M}$ for all ${n}$.

With a bit more effort one can prove that this property of preserving absolute convergence is equivalent to being a bounded sequence. Indeed, if ${\{b_n\}}$ is unbounded, then for every ${k}$ there is ${n_k}$ such that ${|b_{n_k}|\ge 2^k}$. We can ensure ${n_k > n_{k-1}}$ since there are infinitely many candidates for ${n_k}$. Define ${a_n=2^{-k}}$ if ${n = n_k}$ for some ${k}$, and ${a_n=0}$ otherwise. Then ${\sum a_n}$ converges but ${\sum a_nb_n}$ diverges because its terms do not approach zero.

What if we drop “absolutely”? Let’s say that a sequence ${\{b_n\}}$ preserves convergence of series if for every convergent series ${\sum a_n}$, the series ${\sum a_n b_n}$ also converges. Being bounded doesn’t imply this property: for example, ${b_n=(-1)^n}$ does not preserve convergence of the series ${\sum (-1)^n/n}$.

Theorem. A sequence ${\{b_n\}}$ preserves convergence of series if and only if it has bounded variation, meaning ${\sum |b_n-b_{n+1}| }$ converges.

For brevity, let’s say that ${\{b_n\}}$ is BV. Every bounded monotone sequence is BV because the sum ${\sum |b_n-b_{n+1}| }$ telescopes. On the other hand, ${(-1)^n}$ is not BV, and neither is ${(-1)^n/n}$. But ${(-1)^n/n^p}$ is for ${p>1}$. The following lemma describes the structure of BV sequences.

Lemma 1. A sequence ${\{b_n\}}$ is BV if and only if there are two increasing bounded sequences ${\{c_n\}}$ and ${\{d_n\}}$ such that ${b_n=c_n-d_n}$ for all ${n}$.

Proof. If such ${c_n,d_n}$ exist, then by the triangle inequality $\displaystyle \sum_{n=1}^N |b_n-b_{n+1}| = \sum_{n=1}^N (|c_n-c_{n +1}| + |d_{n+1}-d_n|) = \sum_{n=1}^N (c_{n+1}-c_n) + \sum_{n=1}^N (d_{n+1}-d_n)$ and the latter sums telescope to ${c_{N+1}-c_1 + d_{N+1}-d_1}$ which has a limit as ${N\rightarrow\infty}$ since bounded monotone sequences converge.

Conversely, suppose ${\{b_n\}}$ is BV. Let ${c_n = \sum_{k=1}^{n-1}|b_k-b_{k+1}|}$, understanding that ${c_1=0}$. By construction, the sequence ${\{c_n\}}$ is increasing and bounded. Also let ${d_n=c_n-b_n}$; as a difference of bounded sequences, this is bounded too. Finally,

$\displaystyle d_{n+1}-d_n = c_{n+1} -c_n + b_n - b_{n+1} = |b_n-b_{n+1}|+ b_n - b_{n+1} \ge 0$

which shows that ${\{d_n\}}$ is increasing.

To construct a suitable example where ${\sum a_nb_n}$ diverges, we need another lemma.

Lemma 2. If a series of nonnegative terms ${\sum A_n}$ diverges, then there is a sequence ${c_n\rightarrow 0}$ such that the series ${\sum c_n A_n}$ still diverges.

Proof. Let ${s_n = A_1+\dots+A_n}$ (partial sums); then ${A_n=s_n-s_{n-1}}$. The sequence ${\sqrt{s_n}}$ tends to infinity, but slower than ${s_n}$ itself. Let ${c_n=1/(\sqrt{s_n}+\sqrt{s_{n-1}})}$, so that ${c_nA_n = \sqrt{s_n}-\sqrt{s_{n-1}}}$, and we are done: the partial sums of ${\sum c_nA_n}$ telescope to ${\sqrt{s_n}}$.

Proof of the theorem, Sufficiency part. Suppose ${\{b_n\}}$ is BV. Using Lemma 1, write ${b_n=c_n-d_n}$. Since ${a_nb_n = a_nc_n - a_n d_n}$, it suffices to prove that ${\sum a_nc_n}$ and ${\sum a_nd_n}$ converge. Consider the first one; the proof for the other is the same. Let ${L=\lim c_n}$ and write ${a_nc_n = La_n - a_n(L-c_n)}$. Here ${\sum La_n}$ converges as a constant multiple of ${\sum a_n}$. Also, ${\sum a_n(L-c_n)}$ converges by the Dirichlet test: the partial sums of ${\sum a_n}$ are bounded, and ${L-c_n}$ decreases to zero.

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

Since ${\sum_{n=1}^\infty |b_n-b_{n+1}|}$ diverges, by Lemma 2 there exists ${\{c_n\} }$ such that ${c_n\rightarrow 0}$ and ${\sum_{n=1}^\infty c_n|b_n-b_{n+1}|}$ diverges. Let ${d_n}$ be such that ${d_n(b_n-b_{n+1}) = c_n|b_n-b_{n+1}|}$; that is, ${d_n}$ differs from ${c_n}$ only by sign. In particular, ${d_n\rightarrow 0}$. Summation by parts yields

$\displaystyle { \sum_{n=1}^N d_n(b_n-b_{n+1}) = \sum_{n=2}^N (d_{n}-d_{n-1})b_n + d_1b_1-d_Nb_{N+1} }$

As ${N\rightarrow\infty}$, the left hand side of does not have a limit since ${\sum d_n(b_n-b_{n+1})}$ diverges. On the other hand, ${d_1b_1-d_Nb_{N+1}\rightarrow d_1b_1}$ since ${d_N\rightarrow 0}$ while ${b_{N+1}}$ stays bounded. Therefore, ${\lim_{N\rightarrow\infty} \sum_{n=2}^N (d_{n}-d_{n-1})b_n}$ does not exist.

Let ${a_n= d_n-d_{n-1}}$. The series ${\sum a_n}$ converges (by telescoping, since ${\lim_{n\rightarrow\infty} d_n}$ exists) but ${\sum a_nb_n}$ diverges, as shown above.

In terms of functional analysis the preservation of absolute convergence is essentially the statement that ${(\ell_1)^* = \ell_\infty}$. Notably, the ${\ell_\infty}$ norm of ${\{b_n\}}$, i.e., ${\sup |b_n|}$, is the number that controls by how much ${\sum |a_nb_n|}$ can exceed ${\sum |a_n|}$.

I don’t have a similar quantitative statement for the case of convergence. The BV space has a natural norm too, ${\sum |b_n-b_{n-1}|}$ (interpreting ${b_0}$ as ${0}$), but it’s not obvious how to relate this norm to the values of the sums ${\sum a_n}$ and ${\sum a_nb_n}$.

## 1 thought on “Multipliers preserving series convergence”

1. 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 $(x_1, x_2, x_3, ...)$ to $(x_1, x_2-x_1, x_3-x_2, ... )$ so that the paring is given by

$x_1b_1 + \sum_{j=2}^\infty (x_j-x_{j-1})b_j = \sum_{j=1}^\infty x_j (b_j - b_{j+1})$