Matrix Geometric Series For scalar geometric series, we know
$$ \sum_{k=0}^{\infty} x^k  = \dfrac{1}{1-x} \text{ and } \sum_{k=0}^{\infty} kx^{k-1} = \dfrac{1}{(1-x)^2}\,.$$
Does the second one extend to square matrices? We know for $A$ being a $n \times n$ square matrix and $\|A\| < 1$, $\sum_{k=0}^{\infty} A^k = (I-A)^{-1}$. Does the following hold?
$$\sum_{k=0}^{\infty} k A^{k-1} = (I-A)^{-2} $$
 A: Hint. 
\begin{align}
(I-A)^{-2}
=&
\big[(I-A)^{-1}\big]^2
\\
=&
\big[\sum_{k=0}^{\infty} A^k\big]^2
\\
=&
\big[A^0+A^1+A^2+\ldots +A^{k_0-1}+A^{k_0}+\sum_{k=k_0+1}^{\infty} A^k\big]^2
\\
\end{align}
A: We know that $$\sum_{k=0}^n x^k = \frac{x^{n+1} - 1}{x-1}$$
so differentiating that gives $$\sum_{k=1}^{n} kx^{k-1} = \frac{nx^{n+1} - (n+1)x^n+1}{(x-1)^2}$$
or $$(x-1)^2\left(\sum_{k=1}^{n} kx^{k-1}\right) = nx^{n+1} - (n+1)x^n+1$$
The evaluation map $\mathbb{C}[x] \to M_n(\mathbb{C}) : p \mapsto p(A)$ is an algebra homomorphism so we get
$$(A-I)^2\left(\sum_{k=1}^{n} kA^{k-1}\right) = nA^{n+1} - (n+1)A^n+I$$
The series $\sum_{k=1}^{\infty} kx^{k-1}$ converges for $|x| < 1$ so if $\|A\| < 1$ the series
$$\sum_{k=1}^{n} k\|A^{k-1}\| = \sum_{k=1}^{n} k\|A\|^{k-1}$$
also converges, and hence $\sum_{k=1}^{\infty} kA^{k-1}$ exists. On the other hand $$\|nA^{n+1} - (n+1)A^n\| \le n\|A\|^{n+1} + (n+1)\|A\|^n \xrightarrow{n\to\infty} 0$$
Therefore, letting $n\to\infty$ in the above relation gives
$$(A-I)^2\left(\sum_{k=1}^{\infty} kA^{k-1}\right) = I$$
so $(A-I)^{-2} = \sum_{k=1}^{\infty} kA^{k-1}$.
