Prove a sequence is a Cauchy and thus convergent

Question

Suppose that $0<\alpha<1$ and that $\{ x_n\}$ is a sequence which satisfies $$|x_{n+1}-x_n| \le \alpha^n$$ $$n= 1,2,....$$

Prove that $\{x_n\}$ is a Cauchy sequence and thus converges.

Give an example of a sequence $\{ y_n\}$ s.t $y_n \to \infty $ but $$|y_{n+1}-y_n| \to 0$$ as $n \to \infty$

So here's my take so far, in order to understand from the beginning the definition of Cauchy is $\{v_n\}_n$ is a Cauchy sequence if for all $\varepsilon>0$ there exists $N\in \Bbb N$ such that for all natural numbers $n,m\geq N$: $|v_n-v_m|<\varepsilon$.

and I said if $n > m$ $$|x_n-x_m| \le |x_n-x_{n-1}|+|x_{n-1}-x_{n-2}|....+|x_{m+1}-x_m|$$ $$\le \alpha^{n-1}+\alpha^{n-2}+.............+\alpha^{m}$$ $$= 1-\frac{\alpha^{n-m}}{1-\alpha}$$

but from here I'm not quite convinced how I could process further...

Could i get some help?

Answer 1

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

Answer 2

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 $n\rightarrow \infty$