Rigorous way of proving infimum of a sequence

Question

Would someone be able to provide me with a rigorous way of proving that inf(1/sqrt(n)) = 0. I have already proven that it exists by the completeness property.

Thank you!

Answer 1

$A:= ${$1/√n$ | $n\in \mathbb{Z+}$}.

$ \inf(A)$ exists and $\inf(A) \ge 0.$

(Since $0$ is a lower bound.)

Assume $\inf(A) =a >0.$

Choose $n_0 > 1/a^2.$ (Archimedes)

Then

$0 < 1/√n_0 < a.$

$\rightarrow$

$a(>0)$ is not a lower bound of $A.$

Hence

$ \inf(A)= 0.$

Answer 2

Perhaps Archimedean principle helps. For every $\epsilon>0$, find some $N\in{\bf{N}}$ such that $\dfrac{1}{N}<\epsilon.$, then for all $n\geq N^{2}$, we have $\dfrac{1}{\sqrt{n}}\leq\dfrac{1}{N}<\epsilon$. Since any such $n$ satsifies $\dfrac{1}{\sqrt{n}}\in\left\{\dfrac{1}{\sqrt{n}}: n\in{\bf{N}}\right\}$, so $\inf\left\{\dfrac{1}{\sqrt{n}}: n\in{\bf{N}}\right\}<\epsilon$. This is true for all $\epsilon>0$ and the infimum is nonnegative, so it is zero.