If $\sum\frac{1}{a_n}$ diverges, then $\sum\frac{1}{n\max_{1\leq k\leq n}\frac{a_{n+1} - a_k}{n-k+1}}$ diverges.

Question

Let $(a_n)$ be a strictly increasing sequence of positive real numbers, and denote $\Delta a_n:= a_{n+1} - a_n.$

We know that if $\displaystyle\sum_{n\in \mathbb{N}} \frac{1}{a_n}$ diverges, then $\displaystyle\sum_{n\in \mathbb{N}} \frac{1}{n\Delta a_n}$ diverges. $\quad (*)$

Next, for each $n\in\mathbb{N},$ define $b_n:= \displaystyle\max_{1\leq k\leq n} \frac{\Delta a_k + \Delta a_{k+1} + \ldots + \Delta a_n}{n-k+1} = \displaystyle\max_{1\leq k\leq n} \frac{a_{n+1} - a_k}{n-k+1}. $

Proposition: If $\displaystyle\sum_{n\in \mathbb{N}} \frac{1}{a_n}$ diverges, then $\displaystyle\sum_{n\in \mathbb{N}} \frac{1}{n b_n}$ diverges.

Comments:

• Since $b_n \geq \Delta a_n,$ if true, this proposition would be an improvement over, i.e. stronger than, $(*).$

• If $\Delta a_n$ is increasing, then $b_n = \Delta a_n,$ so the proposition reduces to problem $(*),$ which is true.

• A possible counter-example might be something like $A = \displaystyle\bigcup_{n\in\mathbb{N}} \left(\left[2^{2^n-1}, 2^{2^n}\right] \cap \mathbb{N} \right),$ but I am not sure of this.