Proving a contraction mapping is a Cauchy sequence 
Let $\phi(x):[a,b]\rightarrow [a,b]$ be a continuous function. Show that if $\phi(x)$ is a contraction mapping on $[a,b]$ then the sequence $\{x^{(k)}\}$ defined by $x^{(k+1)} = \phi(x^{(k)})$ is a Cauchy sequence. 

Attempted solution - Since $\phi(x)$ is a contraction mapping we have $$|x^{(k+1)} - x^{(k)}| = |\phi(x^{(k)}) - \phi(x^{(k-1)})|\leq L|x^{(k)} - x^{(k-1)}|$$ Applying this idea repeatedly we get $$|x^{(k+1)} - x^{(k)}|\leq L^k|x^{(1)} - x^{(0)}|$$ Now consider the term that must be bounded in order to be a Cauchy sequence \begin{align*}
|x^{(m)} - x^{(m+n)}| &= |(x^{(m)} - x^{(m+1)}) + (x^{(m+1)} - x^{(m+2)}) + \ldots + (x^{(m+n-1)} - x^{(m+n)})|\\
&\leq |(x^{(m)} - x^{(m+1)})| + |(x^{(m+1)} - x^{(m+2)})| + \ldots + |(x^{(m+n-1)} - x^{(m+n)})|\\
&\leq (L^m + L^{m+1} + \ldots + L^{m+n-1})|x^{(1)} - x^{(0)}|
\end{align*}
I am not sure how to proceed and show that for some $M$ we can get this inequality to be less than some $\epsilon$.
Any suggestions is greatly appreciated.
 A: You're on the right track. Note that since $0<L<1$, we have
$$ L^m+L^{m+1}+\dots+L^{m+n-1}\leq \sum_{k=m}^{\infty}L^k=\frac{L^m}{1-L} $$
which can be made as small as we like by choosing $m$ large enough.
A: You are very close to a complete proof. All you need is the "final step". 
Here is your proof, completed with the "final step" (in details).

Let $\phi(x):[a,b]\rightarrow [a,b]$ be a continuous function. Show that if $\phi(x)$ is a contraction mapping on $[a,b]$ then the sequence $\{x^{(k)}\}$ defined by $x^{(k+1)} = \phi(x^{(k)})$ is a Cauchy sequence. 

Proof - Since $\phi(x)$ is a contraction mapping we have 
$$|x^{(k+1)} - x^{(k)}| = |\phi(x^{(k)}) - \phi(x^{(k-1)})|\leq L|x^{(k)} - x^{(k-1)}|$$ 
where $0\leq L< 1$.
It folows by induction (that is, applying this idea repeatedly)  we get $$|x^{(k+1)} - x^{(k)}|\leq L^k|x^{(1)} - x^{(0)}|$$ Now consider the term that must be bounded in order to be a Cauchy sequence \begin{align*}
|x^{(m)} - x^{(m+n)}| &= |(x^{(m)} - x^{(m+1)}) + (x^{(m+1)} - x^{(m+2)}) + \ldots + (x^{(m+n-1)} - x^{(m+n)})|\\
&\leq |(x^{(m)} - x^{(m+1)})| + |(x^{(m+1)} - x^{(m+2)})| + \ldots + |(x^{(m+n-1)} - x^{(m+n)})|\\
&\leq (L^m + L^{m+1} + \ldots + L^{m+n-1})|x^{(1)} - x^{(0)}|= \\
& =\left( \sum_{k=m}^{m+n-1}L^k \right) |x^{(1)} - x^{(0)}| \leq 
 \left( \sum_{k=m}^{\infty}L^k \right) |x^{(1)} - x^{(0)}| = \\
& = \frac{L^m}{1-L}|x^{(1)} - x^{(0)}| 
\end{align*}
Given $\varepsilon >0$, since $0\leq L <1$, there is $M\in \mathbb{N}$ such that for all $m>M$ and all $n \in \mathbb{N}$, 
$$|x^{(m)} - x^{(m+n)}| \leq  \frac{L^m}{1-L}|x^{(1)} - x^{(0)}| \leq \varepsilon$$
So the sequence $\{x^{(k)}\}$ is a Cauchy sequence.
