Proposition: Let $(a_n)_{n\in\mathbb{N}}\subset \mathbb{N}$ be a strictly increasing sequence of positive integers that is large i.e. $\displaystyle\sum_{n\in \mathbb{N}} \frac{1}{a_n}$ diverges. Let $\alpha > 0.$ Then there is a convex subsequence $(a_{k_n})_{n\in \mathbb{N}}$ of $(a_n)_{n\in \mathbb{N}}$ such that $\displaystyle\sum_{n\in\mathbb{N}} \frac{1}{a_{k_n}}\ = \alpha.$
Convex means $$ a_2 - a_1 \leq a_3 - a_2 \leq a_4 - a_3 \leq \ldots . $$
I want to check if the following sequence is a counter-example:
$a_1 = 9, a_2 = 10,\ a_3 = 90, a_4 = 91, \ldots, a_{13} = 100,\ a_{14} = 900, a_{15} = 901, \ldots, a_{114} = 1000,\ a_{115} = 9000,\ldots .$
We have: $\displaystyle\sum \frac{1}{a_n} \approx \left(\ln \frac{10}{9} \right)\cdot \infty \approx \left(\frac{1}{9}\right)\cdot\infty = \infty.$
But the maximum possible value of $\displaystyle\sum_{n\in\mathbb{N}} \frac{1}{a_{k_n}}\ $ if $(a_{k_n})$ is convex, is $\ \frac{1}{9} + \frac{1}{10} + \frac{1}{90} + \frac{1}{900} + \frac{1}{9000} + \ldots = \frac{1}{10} + \frac{1}{9} \left( 1 + \frac{1}{10} + \frac{1}{100} + \frac{1}{1000} + \ldots \right) = \frac{1}{10} + \frac{10}{81} = 0.223\ldots. $
So no convex subsequence of this $(a_n)$ can sum to $0.225,$ for example.
I would like to check if my counter-example is correct for the reasoning I have provided.
Yes, your counterexample is good. However, as a nitpick, rather than $$\displaystyle\sum \frac{1}{a_n} \approx \left(\ln \frac{10}{9} \right)\cdot \infty \approx \left(\frac{1}{9}\right)\cdot\infty = \infty$$
I would prefer to write something to the effect of $$\sum_{k=1}^{10^n}\frac1{a_k}>\sum_{k=1}^n\frac1{10}=\frac n{10}\to\infty\;\text{as}\;n\to\infty.$$