Prove that the limit is zero

Question

Prove that $\lim_{n \to \infty} \frac {1}{\sqrt n}=0$.

Attempy: $\forall \varepsilon>0 $, we have to find $M\in N$ such that $|\frac {1}{\sqrt n}-0|<\varepsilon$ for $n \ge M$.

Let $\varepsilon > \frac 1{\sqrt M}$. We can do this since $M \in N$, and note that $n\ge M \rightarrow \sqrt n \ge \sqrt M \rightarrow 1/\sqrt n \le 1/\sqrt M.$

Then, for $n \ge M$, we have that $|\frac {1}{\sqrt n}-0| = |\frac {1}{\sqrt n}| = \frac {1}{\sqrt n}$ (because $n\ge M \in N) = \frac 1{\sqrt M} <\varepsilon$.

Therefore, by definition of convergence, $\lim_{n \to \infty} \frac {1}{\sqrt n}=0$.

This is an assignment question, and marking criteria is quite strict. So, could you pick any minor mistake?

Thank you in advance.

Answer 1

Almost there. I suggest writing in this way: Given $\epsilon>0$, by Archimedean Principle, we can choose some $M\in{\bf{N}}$ such that $\dfrac{1}{M}<\epsilon^{2}$, then for all $n\geq M$, we have $\sqrt{n}\geq\sqrt{M}$, so $1/\sqrt{n}\leq 1/\sqrt{M}$, and hence $\left|\dfrac{1}{\sqrt{n}}-0\right|\leq\dfrac{1}{\sqrt{M}}<\sqrt{\epsilon^{2}}=\epsilon$.

Answer 2

We can simplify as follow

$$|\frac {1}{\sqrt n}-0|<\varepsilon \iff \sqrt n>\frac {1}{\varepsilon}$$

then set $M>\frac{1}{\varepsilon^2}$.

Answer 3

Rephrasing:

Need to show:

For $\epsilon >0$ given there exists a $n_0 \in \mathbb{Z^+}$

such that for $n \ge n_0$:

$|1/√n| < \epsilon$.

Archimedes:

There is a $n_0 \in \mathbb{Z^+}$ such that

$n_o \gt 1/(\epsilon)^2$.

For $n \ge n_0$ we have

$|1/√n| \le |1/√n_0| < \epsilon.$