Is $X$ a Banach space, $b\in x$ a vector and $A\in L\left ( x,x \right ) $ an application that checks the condition $\left |A ^{n} \right |< 1$, for all $n\in \mathbb{N}$.
(a) show that the equation $x=Ax+b$ has the only solution $x= \sum_{k=0}^{\infty }\left (A^{k} \right )b$.
thanks for all help
First note, that $\|A \| < 1$ implies that $$\sum_{k=0}^\infty \|A^k\| \le \sum_{k=0}^\infty \|A\|^k = (1- \|A\|)^{-1} < \infty $$ converges. Hence, as $L(X,X)$ is a Banach space, the series $\sum_{k=0}^\infty A^k$ converges in $L(X,X)$. Let's call its limit $B$. Then \begin{align*} B(\mathrm{Id} - A) &= \sum_{k=0}^\infty A^k(\mathrm{Id} - A)\\ &= \sum_{k=0}^\infty A^k - \sum_{k=0}^\infty A^{k+1}\\ &= \mathrm{Id}. \end{align*} Along the same lines, one sees $(\mathrm{Id} - A)B = \mathrm{Id}$, so $B = (\mathrm{Id} - A)^{-1}$. Now note that $$ x = Ax + b \iff (\mathrm{Id} - A)x = b \iff x = (\mathrm{Id} - A)^{-1}b. $$
EDIT: Note that it is sufficient to have $\|A^n\| < 1$ for some $n$, in view of your other question this seems to be meant instead of all $n$. As: If $\|A^n\| < 1$, and $m \in \mathbb N$ is given, we may write $m = kn + \mu$ with $0 \le \mu < n$ and $k \in \mathbb N$, giving $$ \|A^m\| = \|A^{kn + \mu}\| \le \|A^n\|^k \max_{0 \le \mu < n} \|A^\mu\| =: C \|A^n\|^k $$ One now has $$ \sum_{m=0}^\infty \|A^m\| \le \sum_{k=0}^\infty n\|A^n\|^kC = Cn(1 - \|A^n\|)^{-1} < \infty $$ Now continue as above.