Proposition: Suppose $(x_n)$ is a positive real sequence such that $\displaystyle\sum_n x_n$ converges, then there exists $N\in\mathbb{N}$ such that $\ \large{ x_{ \left\lceil \frac{1}{x_N } \right\rceil } } < \frac{1}{N}. $
Attempt $1$: $\ \displaystyle\sum_n x_n$ converges $\implies \neg \left( x_n \geq \frac{1}{n}\ \forall n\in\mathbb{N} \right). $ Therefore $\exists N\ $ such that $x_N < \frac{1}{N},\ \implies \left\lceil \frac{1}{x_N } \right\rceil > N.$
Now what?
I'm also also not making much progress via proof by contradiction.
But it feels like it must be true somehow…
Best Answer
Here is a general counterexample:
Take an increasing sequence $1 = a_0, a_1, \cdots$ of positive integers. Define $$x_n = \begin{cases} \frac{1}{a_{2i}}, a_i < n < a_{i + 1} \\ \frac{1}{a_{\lfloor{i / 3} \rfloor}}, n = a_i. \end{cases}.$$ Then let's check the conclusion does not hold. If $a_i < N < a_{i + 1}$ for some $i$, then we have $$x_{\lfloor 1 / x_{N} \rfloor} = x_{a_{2i}} = \frac{1}{a_{\lfloor 2i / 3 \rfloor}} \geq \frac{1}{a_i} > \frac{1}{N}.$$ If $N = a_i$ for some $i$, then we have $$x_{\lfloor 1 / x_{N} \rfloor} = x_{a_{\lfloor{i / 3} \rfloor}} = \frac{1}{a_{\lfloor{i / 9} \rfloor}} \geq \frac{1}{a_i} = \frac{1}{N}.$$ So the conclusion does not hold. It remains to ensure that the sum converges. We have $$\sum_n x_n \leq \sum_{i = 0}^\infty \frac{a_{i + 1} - a_{i}}{a_{2i}} + \sum_{i = 0}^\infty \frac{1}{a_{\lfloor{i / 3} \rfloor}}$$ Taking $a_i = 2^i$ ensures RHS converges. So the proposition does not hold.