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If $\sum a_n$ converges, then $(a_n)\rightarrow 0$ and $a_n<\dfrac{1}{n}$ for infinitely many $n$.
But can we show that $\exists k\in\mathbb{N}$ such that $\forall n\ge k, ~a_n<\dfrac{1}{n}$ ?

If we want to show that the statement is false, we must find a sequence $(a_n)$ such that $\sum a_n$ converges and $a_n\ge \dfrac{1}{n}$, for infinitely many $n$.
If we add the condition that $(a_n)$ is non-negative, does it change the answer ?

Additionally, if $(a_n)$ is decreasing, then I believe the statement is true.
Proof: Since $(na_n)\rightarrow 0$, $\exists k\in \mathbb{N}$ such that, $\forall n\ge k,$ $$|na_n|<1\implies a_n< \dfrac{1}{n}.$$ Note, $(a_n)$ must be a non-negative sequence in this case since $(a_n)$ is decreasing and $(a_n)\rightarrow 0$.

Sathvik
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  • This is surely wrong. You can (for example) choose $a_n = 1/\sqrt n$ for infinitely many $n$ (and $a_n = 0$ otherwise) in a way that the series converges. – Martin R Oct 25 '24 at 10:42
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    See for example here: https://math.stackexchange.com/q/2084371/42969 – Martin R Oct 25 '24 at 10:47
  • It's only true if $(a_n)$ is nonnegative and decreasing. – Adam Rubinson Oct 25 '24 at 11:12
  • @AdamRubinson $P$ := $(a_n)$ is non-negative and decreasing.

    $~~Q$ := $\sum a_n$ converges $\implies \exists k\in\mathbb{N}$ such that $\forall n\ge k, ~a_n<\dfrac{1}{n}.~~$

    I know that $P\implies Q$, in other words, $P$ only if $Q$. You are saying that $Q$ only if $P$. I hope I got that right.

     – Sathvik Oct 25 '24 at 11:56
  • I mean that if $(a_n)$ is non-negative and decreasing then it is true, due to the argument you give in the question. But if $(a_n)$ is not decreasing or not non-negative, then isn't necessarily true. – Adam Rubinson Oct 25 '24 at 13:10
  • @AdamRubinson Thank you. It is clear. – Sathvik Oct 25 '24 at 13:13

1 Answers1

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No (for the first question).

Set $$a_{n^2} = \frac{1}{n^2}. $$ Clearly we have $a_k \geq 1/k$ infinitely often. Then for non-square $k$ just set $a_k$ to be any other non-negative thing else that converges, even $a_k = 0$.

Part of the intuition here is that there are loads and loads of infinite sets $A \subset \mathbb{N}$ that are 'thin enough' for it to be the case that $$\sum_{n \in A} \frac{1}{n} < \infty.$$

SBK
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