The following is the definition of a pseudo-Cauchy sequence:
A sequence $a_n$ is pseudo Cauchy if $\forall$ $\epsilon > 0$, $\exists \ N \in \mathbb N$ such that whenever $n \ge N$, $|a_{n+1} – a_n| \lt \epsilon$.
I don't know if this is a valid proof or not, but this is how I tackled it:
I noticed that the definition of pseudo-Cauchy is similar to that of Cauchy, which is defined as:
A sequence $a_n$ is Cauchy if $\forall$ $\epsilon > 0$, $\exists \ N \in \mathbb N$ such that whenever $m,n \ge N$, $|a_m – a_n| \lt \epsilon$.
Cauchy seems more liberal in the sense that any elements in the set have to be less than $\epsilon$, whereas the pseudo-Cauchy has a more strict requirement that neighbors and only neighbors must satisfy the requirement. However, I said that if we let $m = n+1$, that is, just choose the other element in the set to be the neighbor, that Pseudo-Cauchy sets actually are Cauchy. It is well known that Cauchy sequences are bounded, therefore pseudo-Cauchy sequences are bounded because they can be moulded into the definition of Cauchy.
Best Answer
Define $a_n = \sum_{i=0}^n \frac{1}{i+1}$ (in the reals).
Then $|a_{n+1} - a_n| = \frac{1}{n+2}$, so the sequence is pseudo-Cauchy.
But it is a divergent sequence, as is well known (harmonic series).
So no, not all pseudo-Cauchy sequences are Cauchy. And this sequence is unbounded.