I am trying to follow a proof at the end of section 3.2 in Hunter and Nachtergaele regarding the convergence of an algorithm to compute square roots. To prove that convergence is exponentially fast, the first step (page 67) is to show that the distance between the $n$-th iteration and the fixed point $x$ is less than or equal to the distance between any two points in $\left( x_{0}, Tx_{0}, T^{2}x_{0}, \ldots, T^{n}x_{0} \right)$.
I have tried expressing $$ \begin{align} \lvert T^{n}x_{0} - x \rvert &\leq \sum_{i=1}^{\infty}\lvert T^{n+i-1}x_{0}-T^{n+i}x_0 \rvert \\ \lvert T^{m}x_{0} - T^{n}x_{0} \rvert &\leq \sum_{i=1}^{n-m}\lvert T^{m+i-1}x_{0}-T^{m+i}x_0 \rvert \end{align} $$
but I can't seem to perform a manipulation that gets me $$ \lvert T^{n}x_{0} - x \rvert \leq \lvert T^{m}x_{0} - T^{n}x_{0} \rvert $$
Beginning with the definition of a contraction mapping $T$ and a fixed point $x$ we have $$ \begin{align} \lvert T^{n}x_{0}-x \rvert &\leq c \lvert T^{n-1}x_{0}-x \rvert \\ &\leq c^{n-m} \lvert T^{m} x_{0} - x \rvert \\ &\leq c^{n-m} \left( \lvert T^{m} x_{0} - T^{n} x_{0} \rvert + \lvert T^{n} x_{0} - x \rvert \right) \text{.} \end{align} $$
Rearranging, $$ \left( \frac{1}{c^{n-m}} - 1 \right) \lvert T^{n}x_{0}-x \rvert \leq \lvert T^{m} x_{0} - T^{n} x_{0} \rvert \text{.} $$
We have $0 < c < 1$ and $n-m > 0$ so we have $\frac{1}{c^{n-m}} - 1 > 0$. In the special case where $$ \frac{1}{c^{n-m}} - 1 \geq 1 \tag{1} $$ we can write $$ \lvert T^{n}x_{0}-x \rvert \leq \left( \frac{1}{c^{n-m}} - 1 \right) \lvert T^{n}x_{0}-x \rvert \leq \lvert T^{m} x_{0} - T^{n} x_{0} \rvert \tag{2} \text{.} $$
In the example I was trying to figure out, we had $c=\frac{1}{2}$ so the condition for (2) to be true was $2^{n-m} - 1 \geq 1$ which is always true since $n > m$.
It turns out (1) (and consequently (2)) is always true when $c\leq\frac{1}{2}$. For $\frac{1}{2} < c < 1$, the condition (1) can be expressed in terms of $n-m$ $$ n-m> - \frac{\log 2 }{\log c} \text{.} $$
For example, if $T$ contracts "slowly", $c=\frac{99}{100}$, we can still say $$ \lvert T^{n}x_{0}-x \rvert \leq \lvert T^{m} x_{0} - T^{n} x_{0} \rvert $$ if we restrict ourselves to $n-m \geq 69$.