iteration convergence bounds with norm less than 1 Let $x_{k+1} = Bx_k + c$
where $B$ is $n \times n$ matrix $c$ is a vector.
Assume $\|B\| \le \beta <1$
$\|x_k - x_{k-1}\| \le \varepsilon$ for some $k$
Show that $\| x - x_k\| \le \dfrac{\beta\varepsilon}{1 - \beta}$
Thanks a lot...

Observe that $\|x-x_n\| = \|(Bx+c) - (Bx_{n-1}+c)\| = \|B(x-x_{n-1})\| \le \|B\|\|x-x_{n-1}\|$. So, recursively, we get $\|x-x_n\|\le\|B\|^n\|x-x_0\|$. Since $\|B\|<1$, we see that $\lim_{n\to\infty}x_n=x$. Now, by triangle inequality, we have
\begin{equation}
\|x-x_k\|\le \|x-x_{k+n}\| + \sum_{i=1}^n \|x_{k+i}-x_{k+i-1}\|.\tag{1}
\end{equation}
Yet
$$
\sum_{i=1}^n \|x_{k+i}-x_{k+i-1}\|
= \sum_{i=1}^n \|B^i(x_k-x_{k-1})\|
\le \sum_{i=1}^n \|B\|^i \epsilon
\le \sum_{i=1}^\infty \beta^i \epsilon
= \frac{\beta\epsilon}{1-\beta}.
$$
Therefore, by letting $n\to\infty$ on the RHS of (1), the result follows.
