Convergence of square series

Question

Let $\sum a_n$ be a series that converges absolutely. Prove that $\sum a_n^2$ also converges.

My attempt:

If $\sum |a_n|$ converges, then $\sum an$ converges and $\lim_{n\to\infty} a_n = 0$. There is an $M$ so that for all $n \geq M$

$$|a_n| < 1 \implies |a_n^2| < |a_n|$$

By the direct comparison test, $\sum |a_n^2|$ converges because $\sum |a_n|$ converges. We may conclude that $\sum a_n^2$ converges.

QED.

Is this proof correct? If so, were the absolute values necessary or could this have been done without the absolute values? Any assistance much appreciated.

Answer 1

Your proof is mostly correct. There is one fairly inconsequential error: $$|a_n| < 1 \implies |a_n^2| < |a_n|.$$ It really should be $$|a_n| < 1 \implies |a_n^2| \color{red}\le |a_n|,$$ as the strict inequality doesn't account for the possibility that $|a_n| = 0$. But, with or without the strict inequality, the comparison test still applies.

Note that the comparison test requires that the terms of both series be non-negative. That is, the comparison test is a kind of squeeze theorem for series, where a series is squeezed between a convergent series and the $0$ series. Indeed, you can generalise the comparison test so that the lower bound need not be the $0$ series.

So, for this reason, we need absolute values for this argument to work. It turns an absolutely convergent series into a non-negative series. As a specific counterexample where we don't have absolute convergence, recall that $$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$$ converges conditionally by the Leibniz Alternating Series test. Its square is the harmonic series, which famously diverges. So, $\sum a_n$ converging doesn't mean $\sum a_n^2$ converges, when conditional convergence is involved.