Pseudo-Cauchy sequence I have never seen this terminology before, so I will provide the given definition.
A Pseudo-Cauchy sequence is : A sequence $(a_n)$ if for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $|a_{n+1} - a_n | \leq \epsilon \space \forall \space n \geq  N$
So then my question is that is a pseudo-cauchy sequence always converging?
 A: Consider the sequence
$$0,1,\frac{1}{2},0,\frac{1}{3},\frac{2}{3},1,\frac{3}{4},\frac{2}{4},\frac{1}{4},0,\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},1,\frac{5}{6},\frac{4}{6},\frac{3}{6},\frac{2}{6},\frac{1}{6},0, \frac{1}{7},\frac{2}{7},\cdots.$$
This is a pseudo-Cauchy sequence, and every real number between $0$ and $1$ is the limit of a subsequence of our sequence. 
A: Sequence $$a_n = \sqrt n$$ is pseudo-Cauchy but divergent. It is monotonic, strictly increasing. but not bounded, and divergent.
Let it through the sine function and we get another pseudo-Cauchy sequence $$b_n = \sin(\sqrt n),$$ which is bounded (oscillating between $-1$ and $1$) but not convergent, either.
A: For the record it doesn't even work if you ask $a_n$ to be bounded. Take $a_n = \sin(\sum_1^n 1/k)$. Since $\sum_1^n 1/k → ∞ $ and also the difference $\left|\sum_1^{n+1} 1/k - \sum_1^n 1/k\right| = \frac{1}{n+1} \to 0$, $a_n$ oscillates between 1 and -1 forever. But as $\sin$ is Lipschitz,
$$|a_{n+1} - a_{n}| \leq \left|\sum_1^{n+1} \frac{1}{k} - \sum_1^n \frac{1}{k} \right| = \frac{1}{n+1} → 0$$
A: Take $$a_n = \sum_{k=1}^n {1\over k}.$$
