Suppose $\{b_n\}_{n=0}^\infty$ is a bounded sequence of real numbers with the property that $$\sum_{k=0}^n b_k^+\sim\sum_{k=0}^n\left|b_k^-\right|\text{ as }n\to\infty$$ and $\{a_n\}_{n=0}^\infty$ is a nonnegative sequence decreasing to $0$. For clarity, $x_n\sim y_n$ as $n\to\infty$ means that we can write $x_n=(1+c_n)y_n$ for some sequence $\{c_n\}$ with $\lim_{n\to\infty}c_n=0$.

Question: Is the series $\sum_{n=0}^\infty a_n b_n$ necessarily convergent?

I've tested this claim numerically with several choices for $b_n$, listed below, and am tempted to say the answer is yes.

  • $b_n = (-1)^n$
  • $b_n=\sin(n)$
  • $b_n=4$ if $n\text{ mod }5=0$, $b_n=-1$ otherwise.
  • Many others like the previous one.

This list doesn't cover more exotic sequences, e.g. those that are chaotic, and so it misses potential counterexamples, hence why I'm seeking help.

(Long) Motivation: for any $\{a_n\}$ satisfying the above conditions, the convergence of the series $\sum_n \sin(n) a_n$ is analogous to the convergence of $\sum_n (-1)^{n-1}a_n$. Since the positive terms and negative terms of $\sin(n)$ are evenly distributed in number and magnitude, the contribution of the positive terms of $\sum_n \sin(n) a_n$ "balances out", so to speak, with the contribution of the negative ones. It's almost as if the net "weight" of the positive terms "equals" that of the negative ones.

I will now make this notion of equal weight precise. Here are some sequences that intuitively satisfy the "equal weight" condition:

  • $\{(-1)^n\}$. Half of the terms of this sequence are positive and half are negative, with each term having an individual magnitude of $1$.
  • $\{\sin(n)\}$. Asymptotically, the quotient of the number of positive terms by the number of terms approaches $1/2$, and likewise for the negative terms. Unlike the previous example, the terms of this sequence fluctuate through almost all numbers between $-1$ and $1$, but because all such numbers are "evenly distributed" by magnitude, the "weight" (that word again!) of the positive terms is equal to that of the negative terms.
  • $(1,1,1,1,-1,-1,-1,-1,1,1,1,1,-1,-1,-1,-1,\dots)$. This sequence oscillates between segments of four ones and segments of four negative ones, so the "weight" of the positive terms is equivalent to that of the negative ones.
  • $\left\{\sin\left(\frac{\pi n}{2}\right)\right\}=(0,1,0,-1,0,1,0,\dots)$. Identical to the first sequence, but with some zeros inserted in between.

These examples suggest that one possibility for defining the "weight" of the positive/negative terms of a sequence $\{b_n\}$ is by seeing if the positive terms "make up half" of the whole sequence, and likewise for the negative terms. This is made precise as follows:

Definition: We say that the positive terms $\{b_n^+\}$ and negative terms $\{b_n^-\}$ of a sequence $\{b_n\}$ are of equal weight provided that $$\lim_{n\to\infty}\frac{\text{# of positive terms of }\{b_k\}\text{ with index up to }n}{n}=\lim_{n\to\infty}\frac{\text{# of negative terms of }\{b_k\}\text{ with index up to }n}{n}=\frac{1}{2}$$

Some thinking makes it clear this is not enough to ensure the convergence of $\sum_n a_n b_n$.

Exercise: think of a counterexample.

Here's the problem:

The convergence of $\sum_n a_n b_n$ depends not only on the distribution of its positive and negative terms, but also the magnitude of each term. For instance, if $\{b_n\}=(100,-1,100,-1,\dots)$, then the positive terms will dominate the negative ones, making the series $\sum_n a_n b_n$ diverge to $\infty$.

Thus, we need a better definition that more accurately captures our intuition for the "weight" of a set of reals. We can take a step in this direction, and one step closer to the condition $\sum_{k=0}^n b_k^+\sim\sum_{k=0}^n\left|b_k^-\right|$, by thinking about we can assign a weight to a finite set of reals.

Consider a finite set $S=\{x_1,\dots,x_n\}$ of non-negative reals. If we identify each real $x_i$ with a box of weight $x_i$, then the total weight of the set $S$ should be $$W(S)=\sum_{i=1}^n x_i$$ We could then say that two finite sets of reals $S_1, S_2$ have equal weight provided that $W(S_1)=W(S_2)$. Our definition of $W$ naturally leads to $$W(S)=\sum_{i=1}^\infty x_i\text{ for a countably infinite set }S\text{ of non-negative reals}$$ but this leads to difficulties in comparing the weight of two countable sets, since one or both of the series may not converge. We can't simply extend our notion of equality to include $\infty=\infty$, for there are many examples of countably infinite sets $S_1, S_2$ that are intuitively of different weights but nevertheless satisfy $W(S_1)=W(S_2)=\infty$. An example: $S_1=\{1,1,1,1,\dots\}$ and $S_2=\{1,2,3,4,\dots\}$, where $S_2$ "clearly" weighs more than $S_1$.

Notice that for finite $S_1$ and $S_2$, $W(S_1)=W(S_2)$ is equivalent to $W(S_1)/W(S_2)=1$ provided that $S_2$ has at least one nonzero element. The nice thing about this realization is that it gives us an avenue for reasonably extending the notion of equal weight to more classes of countably infinite sets than $W(S_1)=W(S_2)$ can. The key is to look at the partial sums instead of the whole series:

Definition: we say that two sets of non-negative reals $\{x_1,x_2,\dots\}$, $\{y_1,y_2,\dots\}$ are of equal weight provided that $$\lim_{n\to\infty}\frac{\sum_{i=1}^n x_i}{\sum_{i=1}^n y_i}=1$$ To account for the cases where $y_n$ is initially zero, we can tweak this definition to the equivalent $$\sum_{i=1}^n x_i\sim\sum_{i=1}^n y_i\text{ as }n\to\infty$$

I have not taken the time to study whether this characterization is independent of the order in which each sum is taken, which seems like a natural thing to expect in this context. It seemed reasonable enough as it was, so I stuck with it. In our case of studying the weight of the positive/negative terms of $b_n$, this leads us to stipulate that $$\sum_{k=0}^n b_k^+\sim\sum_{k=0}^n\left|b_k^-\right|\text{ as }n\to\infty$$ Why do we assume that $\{b_n\}$ is bounded at the beginning? Well, I want this convergence test to work for any $\{a_n\}$ decreasing to $0$, and I've found counterexamples if we allow $\{b_n\}$ to be unbounded:

Take $\{b_n\}=\left\{(-1)^{n-1}n\right\}$ and $\{a_n\}=\{1/n\}$. Then $\{b_n\}$ satisfies the equal weight condition and $\{a_n\}$ decreases to $0$, but $\sum_n a_n b_n =\sum_n (-1)^{n-1}$ diverges.

Interesting Observation: Dirichlet's criterion says that having $\sum_{n=0}^N b_n$ be bounded in absolute value for all $N\in\mathbb N$ is enough to imply the convergence of $\sum_{n=0}^\infty a_n b_n$. It turns out that the boundedness of $\sum_{n=0}^N b_n$ also implies the following: $$\sum_n b_n^+\text{ converges if and only if }\sum_n \left|b_n^-\right|\text{ converges}$$ This is a consequence of $\sum_{n=0}^N b_n=\sum_{n=0}^N b_n^+ -\sum_{n=0}^N |b_n^-|$ and the monotonicity of the sums $\sum_{n=0}^N b_n^+$, $\sum_{n=0}^N |b_n^-|$. In particular, if $\sum_n |b_n^-|$ converges, then $\sum_n b_n^+$ does too and $\sum_n b_n$ converges absolutely. This implies the absolute convergence of $\sum_n a_n b_n$, which is not where the value of Dirichlet's test lies.

Assuming $\sum_n |b_n^-|=\infty$, we can write

\begin{align*} \sum_{n=0}^N b_n &= \sum_{n=0}^N b_n^+ - \sum_{n=0}^N \left|b_n^-\right|\\ &= \sum_{n=0}^N \left|b_n^-\right|\left(\frac{\sum_{n=0}^N b_n^+}{\sum_{n=0}^N \left|b_n^-\right|}-1\right)\text{ for every }N\in\mathbb{N}\text{ sufficiently large} \end{align*}

and, from the boundedness of $\sum_{n=0}^N b_n$,

$$\left|\frac{\sum_{n=0}^N b_n^+}{\sum_{n=0}^N \left|b_n^-\right|}-1\right|\leq\frac{M}{\sum_{n=0}^N \left|b_n^-\right|}\text{ for some }M\in\mathbb{R}$$

which, because we assumed $\lim_{N\to\infty}\sum_{n=0}^N |b_n^-|=\infty$, gives

$$\lim_{N\to\infty}\frac{\sum_{n=0}^N b_n^+}{\sum_{n=0}^N \left|b_n^-\right|}=1$$

This gives more motivation for my question. This limit condition is weaker than the boundedness of $\sum_{n=0}^N b_n$ because the inequality

$$\left|\frac{\sum_{n=0}^N b_n^+}{\sum_{n=0}^N \left|b_n^-\right|}-1\right|\leq\frac{M}{\sum_{n=0}^N \left|b_n^-\right|}$$

goes beyond proving that $\frac{\sum_{n=0}^N b_n^+}{\sum_{n=0}^N \left|b_n^-\right|}$ approaches $1$, telling us the rate at which this occurs. That said, I hope it might still be sufficient if we assume $\{b_n\}$ is bounded.

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    $\begingroup$ Make $b_n$ by interlacing the constant sequence of $1$s, the constant sequence of $-1$s and the harmonic sequence $1/n$. Then do something like $a_n = 1/\ln(2+\lfloor n/3\rfloor)$ $\endgroup$ Sep 18, 2022 at 2:56
  • $\begingroup$ @BrianMoehring this is a nice counterexample, so I'd recommend publishing it as an answer. A question: did you pick this $b_n$ because the terms with $b_n=\pm 1$ cancel out? $\endgroup$ Sep 18, 2022 at 3:23
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    $\begingroup$ Yes. I chose it so that the cancellation was as obvious as possible. I won't be posting it as an answer because Apass.Jack's answer is essentially the same (the only reason I could put out to prefer my own is that showing $\frac1n \sum_{k\leq n} \frac1k \to 0$ is a little easier than $\frac1n \sum_{k\leq n} \frac1{\ln {k+2}} \to 0$). However, I will give an answer to point out the restriction to $b_n$ bounded, without any other restriction, is actually no restriction at all. $\endgroup$ Sep 18, 2022 at 3:51

2 Answers 2


Not necessarily.

Here is an example.

Let $a_{2n}=a_{2n+1}=\frac1{n+2}$, $b_{2n}=1+\frac1{\ln(n+2)}$, $b_{2n+1}=-1$, for all $n\ge0$.

$$\sum_{n=0}^\infty a_n b_n= \sum_{n=0}^\infty (a_{2n} b_{2n} + a_{2n+1}b_{2n+1})=\sum_{n=0}^\infty \frac1{n+2}\frac1{\ln(n+2)}\\ =\sum_{n=2}^\infty \frac1{n}\frac1{\ln n}\ge\int_3^\infty\frac{dx}{x\ln x}=\ln\ln x|_3^\infty=\infty$$

The series is not necessarily convergent even if we also require $\lim\limits_{n\to\infty}nb_n=0$, a condition that is significantly stronger than $\{b_n\}_{n=0}^\infty$ is bounded.

For example, let $a_{2n}=a_{2n+1}=\frac1{\sqrt{\ln n}}$, $b_{2n}=\frac1{n\ln\ln n}+\frac1{n\sqrt{\ln n}}$, $b_{2n+1}=-\frac1{n\ln\ln n}$, for all $n\ge3$.


Note that in the problem as written, you already had enough information to determine the answer, as you wrote here:

Why do we assume that $\{b_n\}$ is bounded at the beginning? Well, I want this convergence test to work for any $\{a_n\}$ decreasing to $0$, and I've found counterexamples if we allow $\{b_n\}$ to be unbounded:

Take $\{b_n\}=\left\{(-1)^{n-1}n\right\}$ and $\{a_n\}=\{1/n\}$. Then $\{b_n\}$ satisfies the equal weight condition and $\{a_n\}$ decreases to $0$, but $\sum_n a_n b_n =\sum_n (-1)^{n-1}$ diverges.

At this point, restricting to $b_n$ bounded explicitly restricts this sequence, but only superficially. The behavior you're trying to avoid is still there. Just replace every term $(-1)^{n-1}n$ with $n$ consecutive terms of $(-1)^{n-1}$ and extend the sequence $a_n$ by making $1/n$ appear $n$ consecutive times as well.

Both the behavior you want (asymptotic parts of $b_n$, decreasing $a_n$) and the behavior you don't (divergence of $\sum a_nb_n$) is still there. In fact, your counterexample just amounts to taking subsequences of this "spread out" version.

  • $\begingroup$ That's a really good insight. I'm now going to investigate what properties on $b_n$ stronger than mere boundedness can fix the test, if any. $\endgroup$ Sep 18, 2022 at 4:05

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