Prove $\sqrt{n}$ to be non-Cauchy

Question

I’m trying to prove that the sequence $(a_n)$ with $a_n=\sqrt{n}$ is not a Cauchy sequence, but I’m quite unsure how to provide a counter example for a general chosen epsilon.

Answer 1

Assume $(a_n)_{n \in \mathbb{N}}$ is Cauchy.

Given $ \epsilon \gt 0$ there is a $n_0$ such that

for $m,n \ge n_0 :$

$|a_m-a_n| \lt \epsilon.$

Choose $m= (k+n_0)^2$ , $n=n_0^2.$

$|a_m-a_n| = |k| \lt \epsilon.$

Choose $k \in \mathbb{Z+}$ such that $k > \epsilon.$ (Archimedes).

Hence not Cauchy.

Answer 2

Hint: $\lim_{n \rightarrow \infty}\sqrt{n} = \infty$

Answer 3

Take $\varepsilon=1$. If $p\in\mathbb N$, then, since $\lim_{n\to\infty}\sqrt n=+\infty$, there is a $m\in\mathbb N$ such that $\sqrt m\geqslant\sqrt p +1$. Let $n=p$. Then $m,n\geqslant p$ and$$\left|\sqrt m-\sqrt n\right|\geqslant 1=\varepsilon.$$

Answer 4

Hint Set $m=4n$

$$\left| \sqrt{4n}-\sqrt{n}\right|=\sqrt{n} $$ can be easily made larger than $\epsilon$.

Answer 5

If $a_n=\sqrt{n}$ then $$ |a_{2n^2}-a_{n^2}|=n\sqrt{2} \not\to 0 \,\,\,\text{ and }\,\,\,\min(2n^2,n^2)=n^2 \to \infty. $$

Answer 6

If $(u_n) $ is Cauchy then we will have

$$\lim_{n\to\infty}(u_{n^2}-u_n)=0$$ since $(u_{n^2}) $ is a subsequence of $(u_n) $.

but

$$u_{n^2}-u_n=\sqrt {n}(\sqrt {n}-1) $$ goes to $+\infty $.

thus $(u_n) $ is not Cauchy.

Answer 7

I think you are overthinking this. As the elements of $\{\sqrt{n}\}$ get further apart from each other as $n$ increase this is clearly not Cauchy.

Finding a counter example is simply a matter of finding $m > n$ so that $\sqrt{m} - \sqrt n > \epsilon$. Or in other words $\sqrt{m} > \sqrt n + \epsilon$. Or in other words $m > (\sqrt n + \epsilon)^2 = n + 2\sqrt{n}\epsilon + \epsilon^2$....

Well, that IS your counter-example. For any $\epsilon > 0$ let and $M$ be any real value. There will always be $n,m$ so that $M < n < n + 2\sqrt{n}*\epsilon + \epsilon^2 < m$ so that $|\sqrt{m} - \sqrt{n}| > \epsilon$.

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And even MORE obvious. In the reals all Cauchy sequence converge. (That's pretty much the definition of the real numbers... more or less). And clearly $\{\sqrt{n}\}$ do not converge.

Answer 8

$b_n:=a_{n^2}$ is a subsequence which is clearly not Cauchy (it is $b_n=n$). Hence, $a_n$ can't be Cauchy.