Suppose $x_n>0$ and $\sum_n x_n$ converges. T/F: $\forall\varepsilon>0,\exists\ N,\ \exists\ M\in [N,N^2]$ s.t. $\frac{x_M}{x_N}<\varepsilon.$

Question

Suppose $x_n > 0\ \forall n\in\mathbb{N},\ $ and $\displaystyle\sum_n x_n $ converges.

Proposition: $\ \forall \varepsilon>0,\ \exists\ N,\ \exists\ M\in [N,N^2]\ $ such that $\frac{x_M}{x_N} < \varepsilon.$

Does replacing $[N,N^2]\ $ by $[1, N^2]$ make things significantly easier? I doubt it.

Note that the proposition would be false if we replaced $[N,N^2]$ with $[N,2N]\ $ or $\ [N,3N]\ $ etc, for then $x_n = \frac{1}{n^2}$ would be a counterexample.

My first thought then was to try $x_n = \frac{1}{n\log^2{n}}\ $ as this is one of the most slowly converging sequences (whose series converges) that I have in my arsenal. But this is not a counter-example, since for all $\varepsilon,\ $ if $\ M=N^2, $ then $ \frac{x_{M}}{{x_N}} = \frac{\log^2{N}}{N\log^2(N^2)} \to 0,\ $ and so there exists a large $N$ that satisfies $\frac{x_M}{x_N} < \varepsilon.$

Of course, I don't know the answer to this question. I'm more interested in the tools and strategies needed to answer such a question. How advanced are these types of problems?

Is the same question but with $\ [N,2^N]\ $ or $\ [1, 2^N]\ $ in place of $[N,N^2]$ simpler to solve, or do some or all these questions have non-standard counterexamples as answers?

Edit: the contrapositive of the original statement is:

Suppose $x_n > 0\ \forall\ n\in\mathbb{N}.\ $ If $\ \exists\ \varepsilon>0\ $ such that $\ n,\ m\in [n, n^2] \implies \frac{x_m}{x_n} \geq \varepsilon,\ $ then $\ \displaystyle\sum_n x_n\ $ diverges. If we assume $\ \exists\ \varepsilon>0\ $ such that $\ n,\ m\in [n, n^2] \implies \frac{x_m}{x_n} \geq \varepsilon,\ $ then we have: $x_1 \geq \varepsilon x_1 (\implies \varepsilon\leq 1),\ \ x_4\geq \varepsilon x_2,\ \ x_3\geq \varepsilon x_2,\ \ x_3\geq \varepsilon x_2,\ \ x_9\geq \varepsilon x_3,\ \ x_8\geq \varepsilon x_3,\ \ \ldots,\ x_4 \geq \varepsilon x_3, \ \ x_3 \geq \varepsilon x_3,\ \ldots.$

Can we prove the sum of such a sequence diverges?

Answer 1

The original statement is true. By contradiction:

• let $f(t) = 2^{2^t}$ (note that $f(0) = 2$, and $f(t+1) = f(t)^2$),
• we can show by induction that $\forall t \ge 0, m \in [f(t), f(t+1)] \implies x_m \ge x_2 \varepsilon^{t+1}$,
• $\ \displaystyle\sum_{n \ge 0} x_n\ = x_0 + x_1 + \sum_{t \ge 0} \sum_{m=f(t)}^{f(t+1)-1} x_m$,
• hence $\ \displaystyle\sum_{n \ge 0} x_n\ \ge x_0 + x_1 + x_2 \sum_{t \ge 0} 2^{2^t} ( 2^{2^t} - 1 ) \varepsilon^{t+1} = \infty$,
• a contradiction.