Proof that $\{n\}$ is a Cauchy Sequence. Where is the fallacy?

Question

We need to show that for every $\varepsilon>0$, $\exists N \in \mathbb R$ such that $n,m > N \implies |n-m|<\varepsilon$.

$|n-m|<n+m$. So, if we make $n+m< \varepsilon$, the result will follow. This will happen if both $n<\varepsilon/2$ and $m<\varepsilon/2$.

We see that $n>-\varepsilon/2$ and $m>-\varepsilon/2$. So, let $N=-\varepsilon/2$. Then $$n,m > N \implies |n-m|<n+m<\varepsilon/2+\varepsilon/2<\varepsilon$$

I know that $\{n\}$ diverges but the steps of the proof seem logical.

Where is the fallacy?

Answer 1

The error in your proof is the assumption $n<\varepsilon/2$ (and similarly for $m$). This condition cannot be satisfied for large $n$, so you cannot deduce $|n-m|<\epsilon$ for all large $m$ and $n$.

Answer 2

No. Let $\epsilon=1/2$. Then for any $N$, just let $n=N+1$ and $m=N+2$ so $|m-n|=1\geq 1/2$.

Also note that $n$ doesn't converge so it can't be Cauchy.

Answer 3

$\{n\}$ is not a Cauchy sequence. To see this, suppose we are given $\epsilon=1$. We are asked to find an $N$ such that $m,n>N\implies |n-m|<1$. Suppose that we claim to have found such an $N$. Then, note that $N+1,N+2>N$, but $|(N+2)-(N+1)|=1\not<1$. Therefore, the sequence is not Cauchy.

Alternatively, you could use the fact that Cauchy sequences in $\mathbb{R}$ converge, but $\{n\}$ diverges to infinity.

Answer 4

The error is that $n$ is large and $\epsilon$ is small.

Answer 5

For every $\varepsilon>0$, there exists $N$ such that $n,m>N$ implies $|n-m|<\varepsilon$

This is indeed the correct definition of the sequence being Cauchy. But you have not clear that it should be spelled out, more precisely,

For every $\varepsilon>0$, there exists $N$ such that, for all integers $n$ and $m$, if $n,m>N$ then $|n-m|<\varepsilon$

I emphasized the key words for all.

Now choose $\varepsilon=1/2$ and suppose $n,m>N$, where $N$ is any positive integer. If $n=\lfloor N\rfloor+3$ and $m=\lfloor N\rfloor+1$, we have $|n-m|=2>1/2$.

Thus the condition “if $n,m>N$ then $|n-m|<\varepsilon$” doesn't hold for every choice of $n$ and $m$ larger than $N$.