# Why is this negation regarding rational Cauchy series true?

$$(x_n)$$ as Cauchy sequence

Suppose the preposition below:

$${(*)}$$ $$|x_n| \geq r$$ holds for all $$n \geq n_0$$ and some rational $$r>0$$.

my professor states that the negation of $${(*)}$$ is:

for each $$N \in \mathbb{N}$$ so there exists $$n_N \in \mathbb{N}$$ such that $$\displaystyle|x_{n_{N}}|<1/k$$

I didn't understand this, why did he used subsequencies?

• I created a new account since I didn't get know how figure this out from anywhere, I'll try respect Sep 16, 2022 at 20:33

For any $$r>0$$, any $$n_0$$, there exists $$n\ge n_0$$ with $$|x_n|.
Assume the negation. Because it quantifies over all (rational) $$r>0$$, for any natural $$k$$ I may set $$r=1/k$$. Let $$n_0=1$$ and $$k=1$$. There exists $$n>n_0$$ - call it $$n_1$$ - with $$|x_{n_1}|. Now let $$n_0=n_1+1$$ and $$k=2$$. There exists an $$n_2\ge n_0=1+n_1>n_1$$, with $$|x_{n_2}|. And so on. I can inductively find $$n_1 with $$|x_{n_k}|<1/k$$ for every natural $$k$$.