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})

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