Why is the sequence $x(n) = \log n$ **not** Cauchy?

Question

I read in the book Applied Analysis by Hunter and Nachtergale that

the sequence $x(n)=\log(n)$ is not Cauchy since $\log(n)\to\infty$

But that seems to be irrelevant to the definition of a Cauchy sequence which I understand is as follows:

A sequence $x(n)$ is said to be Cauchy if for every $\epsilon > 0$ there exists an $N$ such that $\lvert x(m)-x(n)\rvert < \epsilon$ for all $m,n>N$.

This sequence $(\log n)$ seems to meet the definition. So how come it is not considered Cauchy?

Answer 1

It's not clear why you say the sequence $(\log(n))$ "seems" to satisfy the definition of Cauchy sequence, but in fact it does not.

It's trivial from the definition that any Cauchy sequence is bounded; no need to invoke the fact that Cauchy sequences of reals are convergent: Say $(x_n)$ is Cauchy. The definition shows that there exists $N$ such that $|x_n-x_m|<1$ for every $n,m> N$. So in particular $|x_{N+1}-x_n|<1$ for every $n>N$, and now the triangle inequality shows that $$|x_n|<|x_{N+1}|+1\quad(n>N).$$So $x_n$ is bounded.

Answer 2

Because it can not converge. "$x_n$ is a Cauchy sequence if and only if it is convergennt". ($x_n$ is a real sequence).

Answer 3

Hint. Look at the difference $$ \log(2n) - \log(n) . $$

Answer 4

The short answer is that $\mathbb{R}$ is complete so every Cauchy sequence has a limit in $\mathbb{R}$. Thus since $log(n)$ does not tend to a real number it cannot possibly be Cauchy.

To see why your idea doesn't work fix $\epsilon=1$

Note $|log(m)-log(n)| = |log(\frac{m}{n})|$

For any N you give me, pick some $n>N$ then for any $m>en$ we have $|log(m/n)|>1=\epsilon$

Edit: in response to comment on original post, we do not need infinity to "count" - every Cauchy sequence has a finite limit. It is easy to see every Cauchy sequence is bounded since $\exists N \in \mathbb{N}: |x_n-x_N|<1 \forall n>N$ Since $x_N$ fixed the sequence is bounded for $n>N$ and the first N terms are a finite sequence so bounded.

Answer 5

Answering late since there does not seem to be a complete proof for your question. For the sake of contradiction, suppose $(\log(n))_n$ is Cauchy. Let $\epsilon>0$. Then there exists $N \in \mathbb{Z}_+$ such that $|\log(n)-\log(m)|<\epsilon$ for $n \geq m \geq N$. Remark $|\log(n)-\log(m)|=|\log\left(\frac{n}{m}\right)|=\log\left(\frac{n}{m}\right)<\epsilon$ for $n \geq m \geq N$. The elimination of the absolute value is because $\log\left(\frac{n}{m}\right)\geq 0$ for any $n,m \in \mathbb{Z}_+$ with $n \geq m$.

Recall the exponential function is increasing, and hence preserves inequality. It follows $\frac{n}{m}<e^\epsilon$. Now suppose $n>\max(N,me^\epsilon,m)$. It follows $\frac{n}{m}>e^\epsilon$. This is a contradiction. Therefore $(\log(n))_n$ cannot be Cauchy. $\square$