$\newcommand{\nn}{\mathbb N}\newcommand{\abs}[1]{\left| #1 \right|}\newcommand{\d}{\,\mathrm d}$When I was thinking about a problem I thought the following conjecture to be true.

$$\sum_{n \in \nn} a_n \; \text{converges absolutely} \iff \exists \gamma \in (1,\infty): \; \abs{a_n} \leq n^{-\gamma} \; \text{for almost all } n \in \nn$$

My intuition was to think about the absolute convergence of a series as an abstract integral on $\nn$ and compare it the convergence of the integrals of the form $\int_1^\infty x^{-\gamma} \d x$ for $\gamma >1$. However this sounded like a too powerful as it would classify all of the absolute convergent series, which shouldn't have gone unnoticed. The direction $\Leftarrow$ is clear. For the $\Rightarrow$ I gave the following proof:

Let's prove the contraposition of the statement, which reads:

$$\forall \gamma \in (1, \infty)\;\forall n_0 \in \nn, \; \exists n \geq n_0 : \;\; \abs{a_n} \gneqq n^{-\gamma} \implies \sum_{n\in \nn} a_n \; \lnot\,\text{converge absolutely}$$

Notice that if you let finitely many elements out of the series, you don't change the convergence behaviour of the series so without loss of generality assume that:

$$\forall \gamma \in (1,\infty), \;\forall n \in \nn\,:\; \abs{a_n} \gneqq n^{-\gamma}$$

since the above inequality holds for all $\gamma$ we can take the supremum of both sides and get:

$$\forall n \in \nn: \; \abs{a_n}\gneqq n^{-1} \tag 1$$

note however that by comparison test the series $a_n$ diverges since the harmonic series diverges, which shows that the series doesn't converge absolutely, which completes the proof.

Does the above argumentation correct? It seems to be correct even though the result seems quite restrictive.


$\newcommand{\nn}{\mathbb N}\newcommand{\abs}[1]{\left| #1 \right|}\newcommand{\d}{\,\mathrm d}$The argumentation is almost correct, which in return makes the proof completely false. Notice that the element you take out of the series depend on the value of $\gamma$. Meaning that, the series you constructed depends secretly on $\gamma$, which is why you cannot take the supremum of both sides without considering this fact. Let's say that the set, for which the inequality doesn't hold, is $N_\gamma$. Thus you actually have:

$$\forall \gamma \in (1,\infty), \;\forall n \in \nn\setminus N_\gamma\,:\; \abs{a_n} \gneqq n^{-\gamma}$$

If you want to consider the limiting case as $\gamma \to 1$ then you should also consider what happens to the set $N_\gamma$. In fact as you decrease $\gamma$ the set $N_\gamma$ gets larger and larger an it is not so clear, that $\lim_{\gamma \to 1} N_\gamma \neq \nn$ but let's try to prove it anyway (!). We'll try to prove it by contradiction. Define the set:

$$ N:= \{ n \in \nn : \abs{a_n} \geq 1/n \} $$

note that the cardinality of this set is infinite since your second inequality holds for all $n_0 \in\nn$. Then we have:

$$ \sum_{n \in \nn \setminus N} \abs {a_n} + \sum_{n \in N} \abs {a_n} \geq \sum_{n \in N} \abs{a_n} \geq \sum_{n \in N} \frac 1n \overset ? = \infty$$

where I used for the last inequality the property of the set $N$. Note that absolute convergence implies unconditional convergence, so that I can change the order of summation. We'd like to have that $\sum_{n\in N} 1/n = \infty$ which would conclude our proof. Note however that this is far from true. For example consider the set $N$ given as:

$$ N:= \{n \in \nn : n = k^2 \text{ for some } k \in \nn\}$$

as you might know the series $\sum_{n\in \nn} 1/n^2$ converges and choosing the set $N$ like that would give us:

$$ \sum_{n\in N} \frac 1n = \sum_{k \in \nn} \frac 1{k^2} < \infty $$

This consideration provides us with a counterexample. Let's define $a_n$ as follows:

$$a_n = \begin{cases} \frac 1n & \text{if } n = k^2 \text{ for some } k\in \nn \\\\ \frac 1 {n^2} & \text{otherwise}\end{cases} $$

As you can easily see $a_n$ converges absolutely to $2 \cdot \pi^2/6$ because $\sum_{n \in \nn} 1/n^2 = \pi^2/6$. However for all $\gamma \in (1, \infty)$ there exists $n\in \nn$ such that $\abs{a_n} \gneqq n^{-\gamma}$. Since for $k > 1$

$$ a_{k^2} = \frac 1k \gneqq \frac 1{k^2}$$

Thus the statement cannot be true.


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