= Solution
Put $r=\|X\|<1$. By <submultiplicativity of the operator norm>, $\|X^k\|\le r^k$, so the <Neumann series> converges absolutely in the complete finite-dimensional matrix space. Its partial sums satisfy
$$
(I-X)\sum_{k=0}^N X^k=\left(\sum_{k=0}^N X^k\right)(I-X)=I-X^{N+1}.
$$
Taking limits gives \b[the inverse and a quantitative remainder bound]:
$$
\boxed{f(X)=\sum_{k=0}^{\infty}X^k,\qquad
\left\|f(X)-\sum_{k=0}^N X^k\right\|\le\frac{r^{N+1}}{1-r}.}
$$
To establish two actual <Frechet derivatives>, use the <resolvent identity>
$$
f(X+H)-f(X)=f(X+H)Hf(X).
$$
For fixed $X\in U$ and sufficiently small $H$, factor $I-X-H=(I-X)(I-f(X)H)$ and apply the <Neumann series> to the second factor. This proves that $X+H\in U$, that $f(X+H)$ is locally bounded, and that $f(X+H)-f(X)=O(\|H\|)$. Subtracting $f(X)Hf(X)$ in the identity leaves $O(\|H\|^2)$. Thus $Df(X)[H]=f(X)Hf(X)$ is the <Frechet derivative>.
Near zero, $f(X)=I+X+O(\|X\|^2)$, uniformly in the <operator norm>. Hence, as a linear operator in $H$,
$$
Df(X)[H]=H+XH+HX+O(\|X\|^2\|H\|).
$$
This proves differentiability of $Df$ at zero, giving
$$
\boxed{Df(0)[H]=H,\qquad D^2f(0)[H,K]=HK+KH.}
$$
The order of multiplication matters: the second <Frechet derivative> is the symmetric <bilinear map> $HK+KH$, not $2HK$.
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