If $\sum_{n=0}^{\infty}a_n$ converges so $\sum_{n=0}^{\infty}a_n^2 $ too($(a_n)$ sequence of real numbers)

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Let $(a_n)$ a sequence of real numbers. Prove the following or give a counter example:

If $\sum_{n=0}^{\infty}a_n$ converges so $\sum_{n=0}^{\infty}a_n^2 $ converges as well.

I saw that there are some posts concerning the same problem, but i didn't find the case if $(a_n)$ is a sequence of real numbers.

I notice that it doesn't really hold so i found the following counter example:

Let $(a_n)=\frac{(-1)^n}{\sqrt{n+1}}$. By alternating series criterion we can deduce that the series $\sum_{n=0}^{\infty}a_n$ converges. But, $a_n^2=\frac{1}{n+1}$ and so $\sum_{n=0}^{\infty}a_n^2$ diverges. So, if $\sum_{n=0}^{\infty}a_n$ converges, $\sum_{n=0}^{\infty}a_n^2 $ isn't necessarily convergent if $a_n$ is the sequence of real numbers.

I would like to know if this counter-example works for this statement, please.

I would also know if there is a more direct maneer to prove that the statement is wrong (for example passing by Cauchy definition)?