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I found proofs telling that if $\sum a_n$ with $a_n>0$ is convergent (hence absolutely), then $\sum {a_n}^2$ always convergent.

But what happens if we do not suppose that $a_n>0$?

My guess is that as $a_n \to 0$, $\exists n_0$ such that $\forall n \geq n_0, -\frac{1}{2} \leq a_n \leq \frac{1}{2} \implies 0 \leq a_{n}^2 \leq \frac{1}{2}a_n < a_n$ and conclude with comparison test. Is it right ?

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    $\begingroup$ How are you concluding $0\le a_n^2 <a_n$ when $a_n$ might be negative? Note that $x\le y$ does not imply $x^2\le y^2$ when $x$ and $y$ are not assumed to be non-negative: $-1 \le 0$ but $(-1)^2 > 0^2$. $\endgroup$
    – Christoph
    Jan 12, 2021 at 16:06
  • $\begingroup$ A clumsy multiplication, my bad $\endgroup$
    – Kilkik
    Jan 12, 2021 at 16:12

2 Answers 2

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counterexample: $a_n=(-1)^n \cdot \frac{1}{\sqrt{n}}$

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In fact, the only functions $f: \mathbb R \to \mathbb R$ such that $\sum_n f(a_n)$ is convergent whenever $\sum_n a_n$ is convergent are those that are linear in a neighbourhood of $0$. See here

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  • $\begingroup$ What do you mean by linear in a neighborhood of 0 ? $\endgroup$
    – Kilkik
    Jan 12, 2021 at 16:41
  • $\begingroup$ I mean there are some constants $c$ and $\epsilon$ with $\epsilon > 0$, such that $f(x) = c x$ for $-\epsilon < x < \epsilon$. $\endgroup$
    – Robert Israel
    Jan 12, 2021 at 22:52

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