Krdytkyu

Give an example of a sequence in $mathbb{R}$ which has no subsequence which is a Cauchy sequence.

I can find out a sequence that is not a Cauchy sequence such as ${ln(n)}$ once $|ln(n)-ln(n+1)|=0$ but $|ln(n)-ln(2n)|=|ln(frac{1}{2})|>epsilon$
$forall epsilon<ln(frac{1}{2})$

I can still find a subsequence of the type ${ln(2n)}_{2ninmathbb{N}}$ such that $|ln(2n)-ln(2n+1)|=0$

Question:

What should I do to get a sequence that has no Cauchy subsequence?

Thanks in advance!

Take $a_n=n$. Then for any subsequence $n_k$, $|n_k-n_{k-1}|geq 1$. So, it has no Cauchy sub-sequence.

Take a sequence $(a_n)$ such that $a_nto infty$. Then each subsequence $(a_{n_k})$ is such that $a_{n_k}to infty$ and thus can't be Cauchy since necessarily Cauchy sequences are bounded.

Take the sequence $1,2,3,4,ldots$ It has no Cauchy subsequence since the distance between any two distinct terms is at least $1$.

I didn't understand your analysis but sequence $a_n = ln n$ doesn't have Cauchy subsequence since its limit is $infty$

As Foobaz John says, any sequence with $a_ntoinfty$ works, and it's easy to see that $|a_n|toinfty$ is also sufficient. In fact this latter condition is necessary and sufficient. If $|a_n|nottoinfty$, then by definition there is some $c>0$ for which there are infinitely many values $n_i$ such that $|a_{n_i}|<c$ for each $i$. Now by Bolzano-Weierstrass the subsequence $a_{n_i}$ has an infinite subsequence $a_{n_{i_j}}$ which is convergent, and therefore Cauchy.

Bolzano-Weierstrass: Every bounded sequence has a convergent subsequence.

Fact: Every convergent sequence is bounded.

Strategy: Try an unbounded sequence.

Guess: $a_n=n$

Conclusion: (I leave it to you)