Showing $\sum_{k \geq 0} kA^{k}$ converges while $\vert \vert A\vert \vert < 1$

Question

Let $\vert \vert \cdot \vert \vert$ be a matrix norm on $A$ where $\vert \vert A\vert \vert < 1$. Show that $\sum_{k \geq 0} k A^{k}$ converges.

My ideas: Let $m<l$

$1.$ Let $\vert\vert\sum_{k=0}^{l}kA^{k}-\sum_{k=0}^{m}kA^{k}\vert\vert=\vert\vert\sum_{k=m+1}^{l}kA^{k}\vert\vert\leq \sum_{k=m+1}^{l}\vert\vert kA^{k}\vert\vert=\sum_{k=m+1}^{l}\vert k\vert \vert \vert A^{k}\vert\vert\leq \sum_{k=m+1}^{l}\vert k\vert \vert \vert A\vert\vert^{k}$

If I can remove $\vert k \vert$ then I am can show that it is a cauchy sequence and subsequently a convergent sequence.

other ideas: Am I allowed to simply take the derivative of $\sum_{k \geq 0} k A^{k}$, but how would I then be able to compare $\sum_{k \geq 0} k A^{k}$ and $\sum_{k \geq 0} A^{k}$? Looking for tips.

Answer 1

You could do it like this:

1. $\sum_{k=0}^\infty k\cdot q^k$ converges for every $q$ with $q\in(-1,1)$ (Ratio test)

2. $\sum_{k=0}^\infty\|kA^k\|\leq\sum_{k=0}^\infty k\cdot \|A\|^k<\infty$

3. If $(x_k)$ is a sequence in a Banach space with $\sum_{k=0}^\infty\|x_k\|<\infty$ then $\sum_{k=0}^\infty x_k$ converges