# $\sum a_n$ 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

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.

• Commented Oct 19, 2023 at 16:21

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 $$n$$.

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

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

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 $$\sum_nb_na_{n+1}\leq\sum_k\frac{(k!)^2}{((k+1)!)^2}=\sum_k\frac1{(k+1)^2}<\infty$$.

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.

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.

• for the falsity of the second implication for arbitrary $b_n$, take $a_n$ in pairs that decrease very fast (eg $a_{2n-1}=a_{2n}=n^{-n}, n \ge 1$) and $b_n$ in opposite sign equal pairs so $|a_{2n-1}b_{2n-1}|=|a_{2n}b_{2n}|=1/n$ is the harmonic say, (so $b_{2n-1}=n^{n-1}, b_{2n}=-n^{n-1}$ for the above $a_n$); then $\sum a_nb_n$ converges but the shifted sum has a dominant negative $1/n$ term in each pair so it diverges; for a strictly decreasing $a_n$ just perturb the above a little Commented Oct 24, 2023 at 15:44
• @Conrad Thank you for the example, it's been referenced in the answer. Commented Oct 24, 2023 at 17:31