I've got a question concerning some weird kind of Lipschitz constant function, but it's an introduction course in Mathematics so Lipschitz hasn't passed the course yet. I should be able to prove this just by convergence of series, sequences and showing Cauchy. Here's the question
Introduction (This can be skipped)
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function and $a\in\mathbb{R}$. Define the sequence $(x_n)_{n\geq 0}$ by $x_0=a$ and for $n\geq 0, x_{n+1}=f(x_n)$.
Necessary information
Assume now that the function $f(x)$ has the following property: There is a constant $L\in(0,1)$ such that for all $x,y\in\mathbb{R}$ that following holds: $|f(x)-f(y)|\leq L|x-y|$.
Already Shown / Proven $$|x_{k+1}-x_k|\leq L^k|x_1-x_0|$$ For all $n>m\geq 0$ the following holds: $$|x_n-x_m|\leq\sum_{k=m}^{n-1}L^k|x_1-x_0|$$
Assignment
Show that the sequence $(x_n)_{n\geq 0}$ is convergent.
How far i am
I've split up the right hand side of the inequality. And I see that i have to show that $|x_n-x_m|$ is a Cauchy sequence.
$$|x_n-x_m|\leq\sum_{k=m}^{n-1} L^{k}|x_1-x_0|=\sum_{k=0}^{n-1}L^{k}|x_1-x_0|-\sum_{k=0}^{m-1}L^{k}|x_1-x_0|=c \left(\sum_{k=0}^{n-1}L^{k}-\sum_{k=0}^{m-1}L^{k}\right).$$
I've called $|x_1-x_0|=c$ because it's a constant. I also know that the 2 sums will end up as $\frac{1}{1-L}-\frac{1}{1-L}=0$ since they're geometric series. So the right hand side goes to 0 for n and m taken large enough.
I just don't see how i can formalize the proof. I don't have a clue how i can write down (in the form of a mathematically correct proof) that:
$\forall \epsilon >0$ there exists a natural number $N$, such that $m,n\geq N$ implies $|s_n-s_m|<\epsilon.$
Hint: For $n > m$, write $|x_n - x_m| \leq L^m \sum_{k=0}^\infty L^k |x_1 - x_0|$ and then use the fact that the summation gives a universal constant (geometric series) and $L^m \to 0$ as $m \to \infty$.