# If $\sum a_n$ is convergent , then $\sum {a_n}^2$ convergent.

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 ?

• 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 $(-1)^2 > 0^2$. Jan 12, 2021 at 16:06
• A clumsy multiplication, my bad Jan 12, 2021 at 16:12

counterexample: $$a_n=(-1)^n \cdot \frac{1}{\sqrt{n}}$$
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
• I mean there are some constants $c$ and $\epsilon$ with $\epsilon > 0$, such that $f(x) = c x$ for $-\epsilon < x < \epsilon$. Jan 12, 2021 at 22:52