A tricky convergence problem

This is a slight simplification of the Math.SE user @Mike's wonderful solution [1] that is rephrased in my own words.

Theorem. Suppose

1. $(a_n)_{n\geq 1}$ is a non-negative, non-increasing sequence,
2. $(f(n))_{n\geq 1}$ is a sequence of positive integers such that $f(n) \to \infty$, and
3. $\sum_{n\geq 1} a_{f(n)}$ converges.

Then $\sum_{n\geq 1} \frac{a_n}{f(n)}$ also converges.

Proof. We partition $\mathbb{N}_1 = \{1, 2, \ldots\}$ into two subsets $A$ and $B$, where

$$\begin{align*} A &= \{ n \in \mathbb{N}_1 : f(n) > n \}, \\ B &= \{ n \in \mathbb{N}_1 : f(n) \leq n \}. \end{align*}$$ Then it suffices to show that both $\sum_{n\in A} \frac{a_n}{f(n)}$ and $\sum_{n\in B} \frac{a_n}{f(n)}$ are finite.

First sum. Since $(a_n)_{n\in\mathbb{N}_1}$ is non-increasing and $1 \leq f(n) \leq n$ for each $n \in B$, we get

$$\sum_{n\in B} \frac{a_n}{f(n)} \leq \sum_{n \in B} a_{f(n)} \lt \infty.$$

Second sum. Now enumerate $f(\mathbb{N}_1)$ as $\{k_1 \lt k_2 \lt \ldots \}$, and define

$$A_i = A \cap [k_i, k_{i+1})$$ for each $i$. Then the following implication holds true: $$\begin{align*} n \in A_i \quad\implies\quad f(n) \gt n \geq k_i \quad\implies\quad f(n) \geq k_{i+1}. \end{align*}$$ So it follows that $$\sum_{n \in A_i} \frac{a_n}{f(n)} \leq \sum_{n \in A_i} \frac{a_{k_i}}{k_{i+1}} = \frac{k_{i+1} - k_i}{k_{i+1}} a_{k_i} \leq a_{k_i}.$$ Summing both sides over $i$, $$\begin{align*} \sum_{n \in A} \frac{a_n}{f(n)} &\leq \sum_{i=1}^{\infty} a_{k_i} \\ &\leq \sum_{i=1}^{\infty} \sum_{n \mathbin{:} f(n) = k_i} a_{f(n)} \\ &= \sum_{n=1}^{\infty} a_{f(n)} \lt \infty. \end{align*}$$ Therefore the desired conclusion follows.

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References.

1. [1] Mike, If $\sum_{n=1}^{\infty}{a_{[f(n)]}}$ converges, then $\sum_{n=1}^{\infty}{\frac{a_n}{f(n)}}$ converges., URL (version: 2022-06-05): https://math.stackexchange.com/q/4464970