Prove that if $\sum_{n=1}^\infty|a_n| < \infty$, then $\{a_n\} \in \ell^2$

I'm having a little trouble getting started with this proof. Here's what I think we need to show:

$$\sum_{n=1}^\infty |a_n| < \infty \implies \sum_{n=1}^\infty a_n^2 < \infty \implies \{a_n\} \in \ell^2.$$ Definitely I know that $$\sum_{n=1}^\infty|a_n| < \infty \implies \sum_{n=1}^\infty a_n < \infty \implies \lim_{n\to\infty}|a_n| = 0 \text{ and } \lim_{n\to\infty}a_n=0.$$

I'm also aware that $\sqrt{a_n^2} = |a_n|$, but I can't exactly find out how to start. I've thought about several methods that I could use to complete this proof such as showing that the sequence of partial sums $$s_n = \sum_{k=1}^\infty a_k^2$$ is bounded, but even that proves to be challenging because I can't quite see any definitive inequalities that arise for any $a_k$. That is, it is not always true that $a_n^2 \le |a_n|$. Can anyone provide some hint on how to get started?

EDIT: Prior to the answer chosen being posted, I actually found an elegant answer to this question:

Given that $\displaystyle \sum_{n=1}^\infty |a_n|$ is convergent, we know that the sequence of partial sums $\displaystyle s_n = \sum_{k=1}^n|a_k|$ is bounded. Therefore there exists $M>0$ such that $|a_n| \le M$ for all $n\in \mathbb N$. So, $$|a_n| \le M \implies |a_n|^2 \le M|a_n|.$$ Thus, $$0 \le \sum_{k=1}^n |a_k|^2 \le \sum_{k=1}^n M|a_k| \le M\sum_{n=1}^\infty|a_n| < \infty \implies \sum_{n=1}^\infty a_n^2 < \infty \implies \{a_n\}\in\ell^2$$

  • $\begingroup$ Eventually, $|a_n|<1$, and then $a_n^2<|a_n|$. $\endgroup$ – David Mitra Nov 9 '16 at 9:31
  • $\begingroup$ Can u help Whats $l^2$? $\endgroup$ – jnyan Nov 9 '16 at 10:03
  • $\begingroup$ @jnyan, the class $\ell^2$ is the class of sequences $\{s_n\}$ such that $\displaystyle \sum_{n=1}^\infty s_n^2 < \infty$ $\endgroup$ – Decaf-Math Nov 9 '16 at 10:13

Since $\sum\limits_{i=1}^\infty a_n < \infty$, we know $a_n$ converges to $0$. So there exists some $N \in \mathbb{N}$ such that $\forall n \geq N$ we have that $|a_n| < 1$. Since $|a_n|$ is a positive number, we multiply both sides of this inequality by $|a_n|$ and obtain $|a_n|^2 < |a_n|$. But $|a_n|^2 = a_n$ hence $a_n^2 < |a_n|$.

Now, may apply this inequality to any $a_n^2$ where $n \geq N$: $$\sum\limits_{i=1}^\infty a_n^2 = \sum\limits_{i=1}^{N-1} a_n^2 + \sum\limits_{i=N}^\infty a_n^2 \leq \sum\limits_{i=1}^{N-1} a_n^2+ \sum\limits_{i=N}^\infty |a_n| \leq \sum\limits_{i=1}^{N-1} a_n^2+ \sum\limits_{i=1}^\infty |a_n|$$ Consider the RHS: the first sum is a finite sum hence it is finite, and the second sum is finite by assumption. Therefore $$\sum\limits_{i=1}^\infty a_n^2 < \infty$$


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