Q:

Quick Proof Verification: Showing Matrix is Nonsingular.

Hello everyone I just want to verify if my proof is correct as I'm very worried that it's not: Thanks so much.
Question: 
Let $A$ be a square matrix and let $\lambda$ be a complex number for which $|\lambda| >\|A\|$, where $\|\cdot \|$ denotes some induced matrix norm.  Show that the matrix $\lambda I-A$ is nonsingular and that
$({\lambda I-A})^{-1}= \frac{1}{\lambda}\sum_{k=0}^{\infty}\frac{1}{\lambda^{k}}A^k$
Proof:
Since $\|A\| \geq 0$ and $|\lambda| >\|A\|$ implies $\lambda$ implies $\lambda\neq 0$
$\|\frac{1}{\lambda}A\|<1$
$\sum_{k=0}^{\infty}\frac{1}{\lambda^{k}}A^k$ converges.
Call $C=\frac{1}{\lambda}\sum_{k=0}^{\infty}\frac{1}{\lambda^{k}}A^k$
Now:
$({\lambda I-A})C={\lambda}C-AC
=\sum_{k=0}^{\infty}\frac{1}{\lambda^{k}}A^k
-\sum_{k=0}^{\infty}\frac{1}{\lambda^{k+1}}A^{k+1}=I$.
Obviously I would also have to show that
$C({\lambda I-A})=I$ as well, but it reduces to the same two series subtracted from each other.
The problem I'm having is understanding why the two series subtracted above=Identity.  I know that that they should., I'm just not sure why exactly.  Any help would be much appreciated.  Thank you.

A:

\begin{align} 
&\sum_{k=0}^\infty \frac{1}{\lambda^k}A^k - \sum_{k=0}^\infty \frac{1}{\lambda^{k+1}}A^{k+1}\\
&= \left(I+ \sum_{k=1}^\infty \frac{1}{\lambda^k}A^k\right) - \sum_{l=1}^\infty \frac{1}{\lambda^{l}}A^{l}\\
&=I+\left( \sum_{k=1}^\infty \frac{1}{\lambda^k}A^k- \sum_{l=1}^\infty \frac{1}{\lambda^{l}}A^{l}\right)\\
\\&=I
\end{align}
where I have let $l=k+1$.

A:

$\sum_{k=0}^{\infty}B^k-\sum_{k=0}^{\infty}B^{k+1}=$
$=I+B+B^2+B^3 +...-(B+B^2+B^3+...)=I$.

A:

$$\sum_{k=0}^\infty\lambda^{-(k+1)}A^{k+1}
=\sum_{k=1}^\infty\lambda^{-k}A^k=-I+\sum_{k=0}^\infty \lambda^{-k}A^k$$
etc. Or
\begin{align}
\sum_{k=0}^\infty \lambda^{-k}A^k-\sum_{k=0}^\infty \lambda^{-(k+1)}A^{k+1}&=
I+\lambda^{-1}A+\lambda^{-2}A^2+\lambda^{-3}A^3+\cdots\\
&{ }-(\lambda^{-1}A+\lambda^{-2}A^2+\lambda^{-3}A^3+\cdots)\\
&=I.
\end{align}

