Be $ 0 < q < 1, (a_n)_{n \in \mathbb{N}} $ a sequence in $\mathbb{R}$ and $ n_0 \in \mathbb{N} $
One has: $ |a_{n+1} -a_n| \leq q |a_n-a_{n-1}|$
for all $n \geq n_o$. Show, that the sequence $(a_n)_{n \in \mathbb{N}}$ convergates.
I tryed to verify, that its a Cauchy-sequence, but i didn't gone very far..
Hint 1:
$|a_{n+1} - a_n| \leq q^{n-1}|a_2 - a_1|$ (why?)
Hint 2:
For $m\leq n\in \mathbb{N}$ $|a_{n+1} - a_{m}| \leq \sum_{k=m}^n|a_{k+1}-a_k|$ by the triangle inequality
Hint 3:
For any $\epsilon>0$, we would like to show that there is an $N$ such that $|a_{n+1} - a_{m}|<\epsilon$ whenever $N<m\leq n$.
Now, for a particular choice of $N$, we have $$ \begin{align} |a_{n+1} - a_{m}| &\leq \sum_{k=m}^n|a_{k+1}-a_k|\\ &\leq \sum_{k=m}^nq^{k-1}|a_2-a_1| \\ &\leq \sum_{k=N+1}^\infty |a_2-a_1| q^{k-1} = |a_2-a_1| q^{N} \sum_{k=0}^\infty q^k\\ &= |a_2-a_1| q^{N} \cdot \frac{1}{1-q} \end{align} $$ You may choose an $N$ so that this is less than $\epsilon$ in order to complete your proof.
