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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.

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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

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  • $\begingroup$ Awesome, many thanks! I think you can replace $\log(n-1)$ even by $\log(n)$. $\endgroup$
    – max_zorn
    Sep 24, 2019 at 15:36
  • $\begingroup$ Sure as actually the modified term is bigger than the corresponding harmonic one but I had the idea using odd terms in mind when writing and then got lazy to edit as it doesn't matter for the problem itself. $\endgroup$
    – Conrad
    Sep 24, 2019 at 17:16

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