# Square summable series that are not summable

Let $$(a_n)$$ be decreasing sequence of positive real numbers such that the square summabiliity assumption $$\sum_{n} a_n^2 < +\infty. \tag{SSA}$$ holds. It is known that in this case $$na_n^2\to 0$$ (see, e.g., https://www.encyclopediaofmath.org/index.php/Series) and hence $$\sqrt{n}a_n\to 0.$$ In turn, this implies (using Cauchy-Schwarz) that $$a_n(a_1+\cdots+a_n)\leq a_n\sqrt{n}(a_1^2+\cdots+a_n^2)\to 0$$ and hence that $$a_n(a_1+\cdots+a_n)\to 0. \tag{*}$$

My question is: What happens if we drop the assumption that $$(a_n)$$ be decreasing? Is (*) still true just assuming the square summability assumption (SSA)? Any reference or counterexample would be much appreciated.

Pick $$a_n = \frac{1}{\log n}, n=2^{2^k}, k \ge 1, a_n =\frac{1}{n}$$ otherwise, Clearly $$\sum a_n^2 < \infty$$ but for every $$n=2^{2^k}, a_1+a_2+...a_n \ge a_1+a_3+...a_{n-1} > \frac{1}{2} \log (n-1)$$ so $$a_n(a_1+a_2+...a_n) > \frac{1}{2}$$ for those infinitely many $$n's$$, so the answer to the question is negative
• Awesome, many thanks! I think you can replace $\log(n-1)$ even by $\log(n)$. Sep 24, 2019 at 15:36