Evaluation of $\lim_{n\to\infty}\sum_{k=1}^{n}kA^{k-1}$ Let
$$\begin{align}
A:&=\begin{pmatrix}{11\over 6}&2\\-1&-1\end{pmatrix}\\
B:&=\lim_{n\to\infty}\sum_{k=1}^{n}kA^{k-1}\\
A^{0}&=I
\end{align}$$
I want to evaluate $B$.
My tries:
$$\begin{align}
A&=\begin{pmatrix}{11\over 6}&2\\-1&-1\end{pmatrix}\\
&={6\over 6}\begin{pmatrix}{11\over 6}&2\\-1&-1\end{pmatrix}\\
&={1\over 6}\begin{pmatrix}11&12\\-6&-6\end{pmatrix}\\
\therefore A^{k-1}&=\left({1\over 6}\right)^{k-1}\begin{pmatrix}11&12\\-6&-6\end{pmatrix}^{k-1}\\
A^{0}&=I\\
A^{1}&={1\over 6}\begin{pmatrix}11&12\\-6&-6\end{pmatrix}\\
A^2&={1\over 36}\begin{pmatrix}49&60\\-30&-36\end{pmatrix}\\
A^3&={1\over 216}\begin{pmatrix}179&228\\-114&-144\end{pmatrix}
\end{align}$$
I coudn't find a regurarity for a power of $A$.
I also tried to use $P^{-1}AP$ but it seems that this kind of approach may not work.
$$
\lambda=\frac{1}{2},~\frac{1}{3}\leftarrow\text{Eigenvalues for}~A
$$
I got $\begin{pmatrix}2&0\\0&1\end{pmatrix}$ after doing elementary roe operations for $\lambda={1\over 2}$ and definitely this set of matrices can't compose a non-zero vector.
I need your help.
 A: We have
$$
(I-2A+A^2)\sum_{k=1}^{n} kA^{k-1} 
= \sum_{k=1}^{n} kA^{k-1}
-2\sum_{k=1}^{n} kA^k
+\sum_{k=1}^{n} kA^{k+1} \\
=I+2A+\sum_{k=3}^{n} kA^{k-1} \\
-2\left(A+\sum_{k=2}^{n-1} kA^k+nA^n\right) \\
+\sum_{k=1}^{n-2} kA^{k+1}+(n-1)A^n+nA^{n+1} \\
=I+2A-2A-2nA^n+(n-1)A^n+nA^{n+1} \\
+\sum_{k=2}^{n-1} (k+1)A^k
-2\sum_{k=2}^{n-1} kA^k
+\sum_{k=2}^{n-1} (k-1)A^k \\
=I-(n+1)A^n+nA^{n+1} +\sum_{k=2}^{n-1} (k+1-2k+k-1)A^k \\
=I-(n+1)A^n+nA^{n+1}
$$
Therefore, if $I-2A+A^2 = (I-A)^2$ is invertible, we have
$$
\sum_{k=1}^{n} kA^{k-1} = (I-A)^{-2}\left(I-(n+1)A^n+nA^{n+1}\right)
$$
As the absolute values of the eigenvalues of $A$ are smaller than $1$ (the eigenvalues are $\frac12$ and $\frac13$), we know $\lim_{n\rightarrow\infty}\left(p(n)A^n\right) = 0$ for any polynomial $p$, see below.
Therefore
$$ \lim_{n\rightarrow\infty}\sum_{k=1}^{n} kA^{k-1}
=(I-A)^{-2}\left(I-0+0\right) = (I-A)^{-2}
$$
The details of $\lim_{n\rightarrow\infty}\left(p(n)A^n\right) =0$
We know that $A$ is diagonalizable and that the eigenvalues are $\frac12$ and $\frac13$. Therefore, we know that there is an invertible matrix $P$ such that
$$ P^{-1}AP = \begin{pmatrix} \frac12 & 0 \\ 0 & \frac13\end{pmatrix}
$$
which means
$$ A = P \begin{pmatrix} \frac12 & 0 \\ 0 & \frac13\end{pmatrix} P^{-1}
$$
Therefore, for a given polynomial $p$:
$$
p(n)A^n = P \begin{pmatrix} \frac{p(n)}{2^n} & 0 \\ 0 & \frac{p(n)}{3^n}\end{pmatrix} P^{-1}
$$
As matrix multiplication is a continuous operation, we can swap the evaluation of the limits and the matrix multiplication:
$$
\lim_{n\rightarrow\infty}\left(p(n)A^n\right)
= P\begin{pmatrix} \lim\limits_{n\rightarrow\infty}\frac{p(n)}{2^n} & 0 \\ 0 & \lim\limits_{n\rightarrow\infty}\frac{p(n)}{3^n}\end{pmatrix} P^{-1}
$$
It is a well known fact that exponential growth is "stronger" than polynomial growth in the sense that the latter divided by the former approaches $0$ as $n$ approaches infinity. Therefore,
$$
\lim\limits_{n\rightarrow\infty}\frac{p(n)}{2^n}
=\lim\limits_{n\rightarrow\infty}\frac{p(n)}{3^n} = 0
$$
for any given polynomial and
$$
\lim\limits_{n\rightarrow\infty}\left(p(n)A^n\right) = 0
$$
A: It's easy to see that the characteristic polynomial of $A$ is $g(x)=(x-\frac12)(x-\frac13).$
Take $f(x)=\sum_{k=1}^nkx^{k-1}.$ Then we can find $h(x),a,b$ such that $$f(x)=h(x)g(x)+ax+b.$$
We can get a linear system
$$\begin{cases}f(\frac12)=\frac12a+b,\\f(\frac13)=\frac13a+b.\end{cases}$$
By Hamilton-Cayley theorem,
$$\lim_{n\to\infty} f(A)=\lim_{n\to\infty}aA+bI= \frac{21}{2}A-\frac{5}{4}I.$$
A: $$\frac{1}{1-x}-1=\sum_{k=1}^\infty x^k$$For $|x|<1$. Differentiating, we know: $$(1-x)^{-2}=\sum_{k=1}^\infty kx^{k-1}$$
Differently put, we know the following formal power series equality: $$\sum_{k=1}^\infty kx^{k-1}(1-x)^2=1$$
This extends to matrices (see the answers to this question) in the sense that: $$\lim_{n\to\infty}\sum_{k=1}^nk A^{k-1}\cdot(1-A)^2=1$$Whenever the series converges. Since the eigenvalues are $1/2,1/3$ and are bounded $<1$, we know convergence holds. So we know $B$ is both a left and right inverse to $(1-A)^2$, hence it is sensible to write: $B=(1-A)^{-2}$.
This is similar to a Von Neumann series, where we use $\|A\|<1$ to prove $(1-A)$ is invertible.
