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

convergence-divergencereal-analysissequences-and-series

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 ?

Best Answer

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

Best Answer

Related Solutions

Show $\sum_{n=1}^\infty \frac {\sqrt a_n}{n}$ is always convergent if $\sum_{n=1}^\infty a_n$ is a convergent series of positive real numbers.

Proof of $\sum a_n$ converges, $a_n \geq 0 \implies \sum a_n^2$ converges.

Related Question