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