Prove a sequence is a Cauchy and thus convergent

You have just have to find an $N$ so that $\alpha^N < \epsilon(1-\alpha)$. Then for $n > m \geq N$, $$\alpha^{n-1} + \cdots + \alpha^m = \alpha^{m}(1 + \alpha + \cdots + \alpha^{n-m-1}) = \alpha^{m} \frac{(1-\alpha^{n-m})}{(1-\alpha)} \leq \frac{\alpha^{m}}{1-\alpha} \leq \frac{\alpha^N}{1-\alpha}$$

An example for the second part can be: $y_n = \sum_{k=1}^n \frac1n$. Then, $|y_{n+1} - y_n|= \frac1{n+1}$, which tends to 0, but $y_n \rightarrow +\infty$ as $n\rightarrow \infty$