Proving sequence converges $\forall n \in \mathbb{N}, |a_{n+1}-a_n| \leq \frac{9}{10}|a_{n}-a_{n-1}|$ 
Given a sequence s.t $\forall n \in \mathbb{N}, |a_{n+1}-a_n| \leq \frac{9}{10}|a_{n}-a_{n-1}|$ prove the sequence converges.

I tried to use cauchy but it didn't work, looking for a hint.
 A: Edited my answer to actually answer the question.
I'll be a bit more explicit than akrihnab, but the same proof strategy.
$\forall i,j \in \mathbb{N}$ with $i>j,$
$$|a_i - a_j| \leq \left( \ \left(\frac{9}{10}\right)^{i-1} + ... + \left(\frac{9}{10}\right)^{j}\ \right)|a_1-a_0| = \left(\frac{1-(9/10)^{i}}{1-9/10} - \frac{1-(9/10)^{j}}{1-9/10}\right)|a_1-a_0| = 10 \left(\left(\frac{9}{10}\right)^{j}-\left(\frac{9}{10}\right)^{i}\right)|a_1-a_0| \leq 10\left(\frac{9}{10}\right)^{j}|a_1-a_0| .$$
Conclusion of above working: $\ \forall n,m \in \mathbb{N}, \ |a_n - a_m| \leq 10|a_1-a_0|\left(\frac{9}{10}\right)^{\min{(n,m)}}.$
Let $\epsilon > 0.$ Then $\exists \ N_0$ such that $\left(\frac{9}{10}\right)^{N_0} < \frac{\epsilon}{10|a_1-a_0|}.\ $ So for $N_0 > \log_{(10/9)}\left(\frac{10|a_1-a_0|}{\epsilon}\right) = \log_{(9/10)}\left(\frac{\epsilon}{10|a_1-a_0|}\right) $:
$\epsilon > 10|a_1-a_0|\left(\frac{9}{10}\right)^{N_0} \geq |a_n-a_m| \ \forall n,m \geq N_0 \implies$ Sequence is Cauchy.
A: We show that the sequence is Cauchy.
Let $|a_m - a_n|$ with $m > n$. Then
$$|a_m - a_n | \leq |a_{m}- a_{m-1}| + \cdots |a_{n+1} - a_n |$$ by triangular inequality. As stated in one of the comments $|a_{k+1} - a_{k}| <= \frac{9^k}{10^k}|a_1 - a_0|$ for any $k$. Since the geometric series $\sum_{k \in \mathbb{N}} \frac{9^k}{10^k}$ converges we conclude that the sequence is Cauchy.
