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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?

$$\sum_{k=0}^{\infty} k A^{k-1} = (I-A)^{-2} $$

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    $\begingroup$ Yes, as long as your norm satisfies $\|AB\|\le\|A\|\|B\|$. $\endgroup$ – Angina Seng Sep 13 '18 at 11:40
  • $\begingroup$ Could you lead me to the result or explain how to get the result? (And yes, the norm is sub-multiplicative) $\endgroup$ – Greenparker Sep 13 '18 at 11:41
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    $\begingroup$ Just multiply $\sum A^k=(I-A)^{-1}$ by itself. $\endgroup$ – Angina Seng Sep 13 '18 at 11:42
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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}

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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 $(A-I)^{-2} = \sum_{k=1}^{\infty} kA^{k-1}$.

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