If $A$ is a contraction, then $(1-\lambda A)^{-1}=\sum_{n=0}^\infty\lambda^nA^n$ for all $|\lambda|<1$ Let $A$ be a self-adjoint contraction (operator norm at most $1$) on a $\mathbb R$-Hilbert space $H$. By contractivity the spectrum $\sigma(A)$ is contained in $[-1,1]$.

Let $\lambda\in[0,1)$. How can we show that $(1-\lambda A)^{-1}=\sum_{n=0}^\infty\lambda^nA^n$?

We may note that $\lambda^{-1}$ is contained in the resolvent set $\rho(A)\supseteq\mathbb R\setminus[-1,1]$ and $1-\lambda A=\lambda(\lambda^{-1}-A)$.
 A: Since
$\Vert A \Vert \le 1, \tag 1$
and
$\lambda \in [0, 1) \Longrightarrow 0 \le \lambda < 1, \tag 2$
the series
$S = \displaystyle \sum_0^\infty \lambda^n A^n \tag 3$
is absolutely and uniformly convergent, viz.
$\left \Vert \displaystyle \sum _0^\infty \lambda^n A^n  \right \Vert \le \displaystyle  \sum _0^\infty \Vert  \lambda^n A^n \Vert$
$= \displaystyle \sum _0^\infty  \lambda^n \Vert  A^n \Vert \le  \sum _0^\infty \lambda^n \Vert  A \Vert^n \le \sum_0^\infty \lambda^n = \dfrac{1}{1 - \lambda} < \infty;  \tag 4$
we further recall the well-known algebraic identity
$(I - \lambda A) \displaystyle \sum_0^m (\lambda A)^n = I - (\lambda A)^m; \tag 5$
when we let $m \to \infty$ in this equation we obtain
$(I - \lambda A) \displaystyle \sum_0^\infty (\lambda A)^n = I; \tag 6$
(6) binds by virtue of the facts that, as has been seen in the above, the sum on the left converges to $S$ whilst $(\lambda A)^m \to 0$, since
$\Vert (\lambda A)^m \Vert = \vert \lambda \vert^m \Vert A^m \Vert \le \lambda^m \Vert A \Vert^m \le \lambda^m \to 0 \; \text{as} \; m \to \infty; \tag 7$
thus $I - \lambda A$ is invertible with
$(I - \lambda A)^{-1} = S. \tag 8$
