Cauchy Sequence Convergence Prove or Give Counterexample Prove or give a counterexample: if a sequence of real numbers $\{x_n\}$ from
$n=1$ to $\infty$
has the property that for all $\epsilon >0$, there exists $N \in \mathbb N$
such that for all $n \ge N$ we have $|x_{n+1} - x_n| < \epsilon$, then $\{x_n\}$ is a convergent sequence. How is this different from the definition of a Cauchy
sequence?
Attempt: For the second part of the problem, I know that it's different from the definition of a Cauchy sequence as it's taking the next part of the sequence and subtracting it from the current part of the sequence.
For the first part, I'm not sure how to go about doing this; originally, I thought I could do something such as $|x_{n+1} - x_n| < \epsilon/2$, in then use the triangle inequality. But I'm not sure we can do that. Thoughts and comments?
And apologies ahead of time for the lack of formatting.
 A: Hint
Try the sequence of harmonic numbers $\displaystyle\sum_{i=1}^n\frac{1}{i}$ and see if it converges.
A: Note that, a sequence ${x_n}$ is Cauchy if
$$ \lim_{n \to \infty} |x_{n+p}-x_n| = 0 \quad \forall p\geq 1. $$
Now, as an example of your case, try $x_n=\ln(n)$.
A: Hint:
In general, every Cauchy sequence is $\mathbb{R}$ is convergent. (Three Steps)


*

*Prove that every Cauchy sequence is bounded.

*Use the Bolzano-Weierstrass Theorem to conclude that it must have a convergent subsequence.

*Show that a Cauchy sequence having a convergent subsequence must itself be convergent.


(The Bolzano-Weierstrass Theorem states that every bounded sequence in $\mathbb{R}$ has a convergent subsequence.)
Note that the sequence given in your question is a special case of a Cauchy sequence.

To clarify the other assumptions:
(1) Harmonic series is NOT Cauchy.
Recall defintion: $\forall \varepsilon >0, \exists N \in\mathbb{N}$ such that $\forall m,n>N,|a_n-a_m|<\varepsilon $
Choose $m=2N$ and $n=N$. Then $$\left | \sum_{i=1}^{2n}\frac{1}{i}  - \sum_{i=1}^{n}\frac{1}{i}  \right |=\sum_{i=1}^{2n}\frac{1}{i}  - \sum_{i=1}^{n}\frac{1}{i} = \sum_{i=n+1}^{2n}\frac{1}{i} = \frac{1}{n+1} +\frac{1}{n+2} + ... + \frac{1}{2n} < \frac{1}{n} +\frac{1}{n} + ... + \frac{1}{n} = n \left (\frac{1}{n}  \right )=1 (=\varepsilon  ) $$
But, remember that $\varepsilon >0$ is chose arbitrarily. So the harmonic series is not Cauchy.
(2) If $a_n$ is Cauchy, then $\lim_{n \to \infty} (x_{n+1}-x_n) = 0$. The converse is NOT true.
Counterexample for the converse:
Take $a_n=\sqrt{n}$.
$\lim_{n \to \infty} (\sqrt{n+1}-\sqrt{n}) = 0$, but $(\sqrt{n})_{n\in\mathbb{n}}$ is not Cauchy.
