I am aware of the fact that $\ell^1(\mathbb{R})$ is a subset of $\ell^2(\mathbb{R})$, but I somehow can't find a very good proof of absolute summability implying square summability across the internet.
Could you kindly verify my following intuition please:
Note that if a infinite series $\sum_{n = 1}^{\infty} a_n < \infty$, then $\lim_{n \to \infty} a_n = 0$
Suppose that $$\sum_{n = 1}^{\infty} |a_n| < \infty, \text{with } a_n \in \mathbb{R}, \forall n$$
By the previous observation, $\exists$ N $\in \mathbb{N}$ s.t. $$\sum_{n = 1}^{\infty} |a_n| = \sum_{n = 1}^{N} |a_n| + \sum_{n = N + 1}^{\infty} |a_n|, \text{ with } |a_n| < 1, \forall n > N$$
And likewise,
$$\sum_{n = 1}^{\infty} a_n^2 = \sum_{n = 1}^{N} a_n^2 + \sum_{n = N + 1}^{\infty} a_n^2, \text{ with } |a_n| < 1, \forall n > N$$
i.e. The infinite sum can be split into a head (which is the finite N-th partial sum) and a tail, thus it suffices to show the convergence of the tail.
By assumption, $$\sum_{n = N + 1}^{\infty} |a_n| < \infty$$
Since $|a_n| < 1, \forall n > N$,
$\implies |a_n|^2 < |a_n|, \forall n > N$
Thus $$\sum_{n = N + 1}^{\infty} |a_n|^2 < \sum_{n = N + 1}^{\infty} |a_n|< \infty$$ by the comparison test.
Hence $$\sum_{n = 1}^{\infty} a_n^2 < \infty$$
This completes the proof. $\tag*{$\blacksquare$}$
Thank you in advance for any comments or suggestions.