Prove that if $\|A\|<1$, then $(I+A)^{-1}=I-A+A^2-A^3+\cdots$. Prove that if $\|A\|&lt;1$, then $(I+A)^{-1}=I-A+A^2-A^3+\cdots$.

I'm not sure how to do this. I know the result for $(I-A)^{-1}$, but that won't help me.
If you know the result for $(I-A)^{-1}$ you can employ it: if $\|A\|&lt;1$, then $\|-A\|=\|A\|&lt;1$; hence $I+A = I -(-A)$ is invertible and $$(I+A)^{-1} = (I-(-A))^{-1} =\sum_{n=0}^\infty (-A)^n = \sum_{n=0}^\infty (-1)^nA^n.$$
Suppose $X=I-A+A^2-\cdots$. Then
$$
\begin{eqnarray}
(I+A)X &amp;=&amp; (I+A)I - (I+A)A + (I+A)A^2 -\cdots \\
&amp;=&amp; (I + A) - (A+A^2) + (A^2+A^3) - \cdots \\
&amp;=&amp; I + (A - A) - (A^2 - A^2) + (A^3 - A^3) - \cdots \\
&amp;=&amp; I
\end{eqnarray}
$$

Since $(I+A)X=I$, we must have $X=(I+A)^{-1}I=(I+A)^{-1}$