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Let $a_n=\sqrt{n}$ with $n\in\mathbb{N}$ and $0<\epsilon\ll1.$

Below I proved that the property $\forall_{\epsilon>0}\exists_{n_0}\forall_{n\geq n_0}(|a_{n+1}-a_n|<\epsilon)$ holds, and that $(a_n)^\infty_{n=1}$is not Cauchy. Posted the proof here for verification, correction and tips on formatting/formalizing!

Proof: The following property must hold:

$-\epsilon<\sqrt{n+1}-\sqrt{n}<\epsilon \iff \sqrt{n}-\epsilon<\sqrt{n+1}<\sqrt{n}+\epsilon \iff$ $n-2\epsilon\sqrt{n}+\epsilon^2<n+1<\epsilon^2+2\epsilon\sqrt{n}+n \iff \epsilon^2-2\epsilon\sqrt{n}<1<\epsilon^2+2\epsilon\sqrt{n} \iff$ $\frac{\epsilon^2}{\sqrt{n}}-2\epsilon<\frac{1}{\sqrt{n}}<\frac{\epsilon^2}{\sqrt{n}}+2\epsilon \iff 0<\frac{1-\epsilon^2}{\sqrt{n}}+2\epsilon<4\epsilon$

We will now proceed to prove the general correctness of both inequalities seperately;

(1) $0<\epsilon\ll1 \iff 2\epsilon>0$

(2) $0<\epsilon\ll1 \iff 0<\epsilon^2<\epsilon\ll 1 \iff 1-\epsilon^2>0$

(3) $n\in\mathbb{N} \iff \forall_n(\sqrt{n}>0)$

So, from (1), (2) and (3), the following property holds: $\forall_{\epsilon>0}\forall_{n\in\mathbb{N}}(\frac{1-\epsilon^2}{\sqrt{n}}+2\epsilon>0)$. This settles the left hand of the inequality. Now we proceed with the right side of the inequality;

$\frac{1-\epsilon^2}{\sqrt{n}}+2\epsilon<4\epsilon \iff \frac{1-\epsilon^2}{\sqrt{n}}<2\epsilon \iff \sqrt{n}>\frac{1-\epsilon^2}{2\epsilon} \iff n>\frac{1-2\epsilon^2+\epsilon^4}{4\epsilon^2}$

We want to find $n_0$ such that $\forall_{\epsilon>0}\exists_{n_0}\forall_{n\geq n_0}(|a_{n+1}-a_n|<\epsilon)$. As $\epsilon$ is constant, such $n_0$ does exist. Let $n_x\equiv \frac{1-2\epsilon^2+\epsilon^4}{4\epsilon^2} \iff n>n_x$. Then, for some $\eta$ we can find $n_0=n_x+\eta$ such that $n\geq n_0$. So, for this found $n_0$, the property $\forall_{\epsilon>0}\exists_{n_0}\forall_{n\geq n_0}(|a_{n+1}-a_n|<\epsilon)$ holds. $\tag*{$\Box$}$

The sequence $(a_n)^\infty_{n=1}$ is not Cauchy.

Proof (1): The following property must hold: $\forall_{\epsilon>0}\exists_{n_0}\forall_{n,m\geq n_0}(|a_n-a_m|<\epsilon)$.

Let $n>m$ without loss of generality, as $n$ and $m$ can be arbitrarily chosen. As $n,m\geq n_0$, the property must also hold as $n,m\to\infty$.

As $n,m\to\infty$ and $n>m, |n-m|=\infty \iff |\sqrt{n}-\sqrt{m}|=\infty $ The property does not hold as $n,m\to\infty$ as of course $\epsilon\ll\infty$. $\tag*{$\Box$}$

Proof (2): In $\mathbb{R}$, every Cauchy-sequence is convergent. Suppose $(a_n)^\infty_{n=1}$ is convergent, then the property $\forall_{\epsilon>0}\exists_{n_0}\forall_{n\geq n_0}\exists_l(|a_n-l|<\epsilon)$ holds. So:

$l-\epsilon<\sqrt{n}<l+\epsilon \iff l^2-2\epsilon l+\epsilon^2<n<l^2+2\epsilon l+\epsilon^2$.

As $l$ is fixed and thus not arbitrarily large, the terms containing $\epsilon$ can be omitted as $\epsilon$ can be chosen arbitrarily small;

As $n\to\infty$, the property $l^2\leq n\leq l^2 \iff n=l^2$ must hold. As $l=\sqrt{n}$ is dependent on $n$, there exists no such unique $l$ such that the property $\forall_{\epsilon>0}\exists_{n_0}\forall_{n\geq n_0}\exists_l(|a_n-l|<\epsilon)$ holds, but by assumption, the series was in fact convergent. This leads to a contradiction, thus the series $(a_n)^\infty_{n=1}$ is not convergent, thus also not Cauchy. $\tag*{$\Box$}$

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  • $\begingroup$ In your final proof you can't fix $m=n+1$ because it has to work for any $m$. However, you can assume that $m>n$ WLOG. $\endgroup$ – Dave Nov 30 '17 at 13:29
  • $\begingroup$ Huh? This sequence is NOT Cauchy! $\endgroup$ – daw Nov 30 '17 at 13:31
  • $\begingroup$ @daw could you explain why not? $\endgroup$ – Marc Nov 30 '17 at 13:31
  • $\begingroup$ Because it has to hold: $$\forall_{\epsilon>0}\exists_{n_0}\forall_{n,m\geq n_0}(|a_n-a_m|<\epsilon)$$ for being a cauchy sequence, but this doesn't hold because for each fixed n you have $\lim_{m\to \infty} |a_m - a_n| = +\infty$ So especially it cannot be arbitrary small for ALL $m,n \ge n_0$ $\endgroup$ – Gono Nov 30 '17 at 13:50
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This sequence is not Cauchy: Fix some $n$, then $$ |a_n - a_m | = |\sqrt n - \sqrt m| \to \infty $$ for $m\to \infty$. Hence if $m$ is sufficiently large compared to $n$, the distance $|a_n-a_m|$ can be made arbitrarily large.

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  • $\begingroup$ Thank you, of course I can see that now with logical reasoning. I've added 2 proofs for the second part now. Could you see if I'm now reasoning correctly? $\endgroup$ – Marc Nov 30 '17 at 14:04
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A Cauchy sequence in $ \mathbb R$ is convergent, your sequence is not convergent!

$a_n \to \infty$ for $n \to \infty$.

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