I don't think so, but I'm not sure as to how to prove this so. A counterexample I can think of is for $a_n = \frac{1}{n}$, here $a_n \ge 0$ for all n and its infinite series when squared does converge. But $a_n$ does not converge, it diverges.
2026-02-23 15:11:13.1771859473
Suppose $a_n \ge 0$ for all n and that $\sum_{n=1}^{\infty} (a_n)^2$ converges, does the $\sum_{n=1}^{\infty} a_n$ also converge?
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No. Consider the counter-example that $a_n = \frac{1}{n}$.