Define an increasing sequence (a_k)_{k \in \mathbb{Z}^+} to be attractive if \sum_{k=1}^\infty \dfrac{1}{a_k} diverges and \sum_{k=1}^\infty \dfrac{1}{a^2_k} converges. Prove that there is an attractive sequence a_k such that a_k\sqrt{k} is also an attractive sequence.
Proposed by Ivan Borsenco.
Solution:
We claim that a_k = \sqrt{k} \log (k+1) do the trick. Indeed, a_k = \sqrt{k} \log (k+1) is obviously increasing for all k \geq 1. Moreover, since \dfrac{1}{k} \leq \dfrac{1}{\sqrt{k}\log (k+1)} for all k \geq 1 and \displaystyle \int_1^\infty \dfrac{1}{x\log^2 (x+1)} \ dx converges we obtain that
\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{\sqrt{k}\log (k+1)} = \infty, \qquad \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{k \log^2 (k+1)} = M < \infty
as a consequence of the Comparison Test and the Integral Test.
Finally, a_k\sqrt{k}=k \log (k+1) is also attractive. As a matter of fact, k \log (k+1) is increasing for all k \geq 1, \displaystyle \int_1^\infty \dfrac{1}{x\log (x+1)} \ dx diverges and \dfrac{1}{k^2 \log^2 (k+1)} \leq \dfrac{1}{k^2} for all k \geq 2, therefore
\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{k\log (k+1)} = \infty, \qquad \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{k^2 \log^2 (k+1)} = N < \infty.
\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{\sqrt{k}\log (k+1)} = \infty, \qquad \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{k \log^2 (k+1)} = M < \infty
as a consequence of the Comparison Test and the Integral Test.
Finally, a_k\sqrt{k}=k \log (k+1) is also attractive. As a matter of fact, k \log (k+1) is increasing for all k \geq 1, \displaystyle \int_1^\infty \dfrac{1}{x\log (x+1)} \ dx diverges and \dfrac{1}{k^2 \log^2 (k+1)} \leq \dfrac{1}{k^2} for all k \geq 2, therefore
\lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{k\log (k+1)} = \infty, \qquad \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{k^2 \log^2 (k+1)} = N < \infty.