$\sum a_n$ diverges $\iff$ for every $(b_n)\subset\mathbb{R}$ s.t. $\sum b_na_n$ diverges, $\sum b_na_{n+1}$ diverges also, and related questions

Question

Suppose $(a_n)_{n\in\mathbb{N}} \subset \mathbb{R}$ is decreasing and $a_n>0\ \forall n\in\mathbb{N}.$ Then which of the following four implications are true, if any?

1. $\ \displaystyle\sum_{n\in\mathbb{N}} a_n$ diverges $\iff$ for every $\ (b_n)_{n\in\mathbb{N}} \subset \mathbb{R}\ $ such that $\ \displaystyle\sum_{n\in\mathbb{N}} b_n a_n$ diverges, $\ \displaystyle\sum_{n\in\mathbb{N}} b_na_{n+1}$ diverges also.

2. $\ \displaystyle\sum_{n\in\mathbb{N}} a_n$ converges $\iff$ for every $\ (b_n)_{n\in\mathbb{N}} \subset \mathbb{R}\ $ such that $\ \displaystyle\sum_{n\in\mathbb{N}} b_n a_n$ converges, $\ \displaystyle\sum_{n\in\mathbb{N}} b_na_{n+1}$ converges also.

I fail to think of counter-examples. To try to help prove the $4$ statements, I suspect the estimates from Cauchy's Condensation test, namely

$$ \sum_{n=1}^{\infty} f(n) \leq \sum_{n=0}^{\infty} 2^n f(2^n) \leq 2\sum_{n=1}^{\infty} f(n) $$

or some version of Schlomilch's generalisation might be useful, but I can't quite see how to use these tools here.

I also wonder how imposing $b_n\geq 0\ \forall\ n\in\mathbb{N}$ in each question would change the result.

Answer 1

There are counterexamples for all implications.

But $\Rightarrow$ in (2) will hold if $b_n\geq0$ or $b_n\leq0$ or $b_n$ is bounded for all $n$.

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Counterexample for $\Leftarrow$ in (1): Consider $a_n=2^{-n}$, we have

$$\sum_nb_na_n\ \text{diverges}\ \ \Rightarrow\ \ \sum_nb_na_{n+1}=\frac12\sum_nb_na_n\ \text{diverges}.$$

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Counterexample for $\Rightarrow$ in (1): With $k\in\mathbb N^+$, consider

\begin{align} &a_n=\frac1{(k!)^2+n/((k+1)!)^2}\ \ \text{if}\ \ (k!)^2\leq n<((k+1)!)^2,\\[.5em] &b_n=(k!)^2\ \ \text{if}\ \ n=((k+1)!)^2\!-\!1\ \ \text{and}\ \ 0\ \ \text{otherwise}. \end{align}

We have

• $a_n>0$ is decreasing and $\sum_na_n\geq\sum_k\frac{((k+1)!)^2-(k!)^2}{(k!)^2+1}=\infty$,
• $\sum_nb_na_n\geq\sum_{k>1}\frac{(k!)^2}{(k!)^2+1}=\infty$, but $\sum_nb_na_{n+1}\leq\sum_k\frac{(k!)^2}{((k+1)!)^2}=\sum_k\frac1{(k+1)^2}<\infty$.

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Counterexample for $\Leftarrow$ in (2): Consider $a_n=1+2^{-n}$, we show that

$$\sum_nb_na_n\ \text{converges}\ \ \Rightarrow\ \ \sum_nb_na_{n+1}\ \text{converges}.$$

Indeed, $\sum_n\!b_na_n$ converges implies $b_n\!\to0$, so $\sum_n\!b_n(a_n\!-\!a_{n+1})=\sum_nb_n2^{-(n+1)}$ is absolutely convergent.

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For $\Rightarrow$ in (2):

If $b_n$ is bounded, then $\sum b_na_{n+1}$ is absolutely convergent.

If $b_n\geq0$ then $\sum_nb_na_{n+1}$ is increasing and bounded above by $\sum_nb_na_n$, so it converges. Similarly for $b_n\leq0$.

As for other situations, see Conrad's comment for the idea to construct a counterexample.