Real numbers sequence convergence So my introduction to analysis professor left this problem as an assignment but I just don't know where to start.
Let $\left( a_n\right)_{n\in \mathbb{N}}$ be a sequence of real numbers and $\alpha \in \mathbb{R}$, $\alpha < 1$. Suppose:
$$ \mid a_{n+1} - a_n \mid \leq \alpha \mid a_n - a_{n-1} \mid \forall n\in \mathbb{N}$$ with $2\leq n$
Show that $\left( a_n\right)_{n\in \mathbb{N}}$ converges
Whenever I see ''$n$'', induction comes to my mind, and maybe Cauchy convertion might be the way to follow, but not exactly sure how to proceed on this one.
Any hints?
Thank you
 A: The trick is to apply your inequality many times (formally, by induction), to get that 
$$
|a_{n+1}-a_n|\leq\alpha^n|a_1-a_0|. 
$$
Then you can use a geometric series to show that the sequence is Cauchy. 
A: You mentioned induction. Use induction to show
$$|a_{n+1} - a_n| \le \alpha^{n-1} |a_2 - a_1|.$$
With this very explicit bound on each increment $|a_{n+1} - a_n|$, you can do a little more work to show this sequence is Cauchy (and thus converges). Specifically use the triangle inequality to write $|a_m - a_n| = |a_m - a_{m-1}| + \cdots + |a_{n+1} - a_n|$, apply the above bound, and use what you know about geometric series to bound $|a_m - a_n|$.
A: Let $b= |a_1 - a_2| \ge 0$
By induction $|A_N - a_{N+1}| \le \alpha^{N-1}*b$.
If $n,m > N$ then $|a_n - a_m| \le |a_n - a_N| + |a_m - a_M|$.
$|a_n - a_N| \le |a_N - a_{N+1}| + |a_{N+1} - a_{N+2}| + .....  + |a_{n-1} - a_n|\le$
$  b*(\alpha^{N-1} + \alpha^{N} + ...... +\alpha^{n-2})$
$= b\sum\limits_{k=N-1}^{n-2}\alpha^k \le$
$b \sum\limits_{k=N-1}^{\infty}\alpha^k = b(\frac 1{1-\alpha} - \frac {1- \alpha^{N-1}}{1- \alpha} ) = b(\frac {\alpha^{N-1}}{1-\alpha})$.
Likewise $|a_m- a_N| < b(\frac {\alpha^{N-1}}{1-\alpha})$.
So $|a_m - a_n| < 2b(\frac {\alpha^{N-1}}{1-\alpha})$.
If for any $\epsilon > 0$ we can find an $N$ so that $2b(\frac {\alpha^{N-1}}{1-\alpha}) < \epsilon$, we will have proven $\{a_i\}$ is Cauchy.
So let $N > \log_{\alpha} (\frac {(1-\alpha)\epsilon}{2b})$ will be such an $N$.
