# prove that : $\sum_{n=0}^\infty |x_n|^2 = +\infty \Rightarrow \sum_{n=0}^\infty |x_n| = +\infty$

i wanted to prove initially that a function is well defined and i concluded that it's enough to prove this statement for $x_n$ a sequence :

$\sum_{n=0}^\infty |x_n|^2 = +\infty \Rightarrow \sum_{n=0}^\infty |x_n| = +\infty$

or :

$\sum_{n=0}^\infty |x_n| < +\infty \Rightarrow \sum_{n=0}^\infty |x_n|^2 < +\infty$

any help ?

If $|x_n|\ge 1$ for infinitely many $n$, then $\sum_{n=1}^\infty|x_n|$ diverges.
If $|x_n|\lt 1$ for all large enough $n$, then for large enough $n$ we have $|x_n|\gt |x_n|^2$, so by Comparison $\sum_{n=1}^\infty |x_n|$ diverges.
Hint: If $\sum a_n < \infty$, then $\displaystyle \lim_{n \to \infty} a_n = 0$. In particular for large $n$, we have $|a_n| \leq 1$. Then
$$|a_n|^2 = |a_n| |a_n| \leq \dots$$
$$\left(\sum|x|\right)^2\geq\sum|x|^2$$ is true for partial sums, so it will be true for the infinite series too. So if $\sum|x|$ is finite, so is$\sum|x|^2$. Your second statement is just the contrapositive of the first, but I think you know that and are just stating it differently.