How to find the limit of a convergent matrix? I'm trying to learn how to show a series of matrices is convergent and find the limit. However my numerical analysis books fail to mention this and I cannot find any relevant material online! Anyway say I had 
$\sum_{k=0}^ \infty A^K$, where A = $\begin{bmatrix}\frac{1}{2} & -2 & -1\\0 & \frac{1}{3} & 0\\ 0 & 0 & -\frac{1}{2}\end{bmatrix}$ 
How would i show that the series of matrrices converges and compute its limit?
Would I use some kind of induction? Some help would be amazing, many thanks!
 A: Hints:


*

*How would you first write the matrix power $A^K$?

*Once you have an expression for $A^K$, you can take the sum to each $a_{i,j}$ in your 3x3. Of course four of the terms are immediately known! Also, two of terms in the diagonal are the same, so you would be finding the sum of four expressions that are sums over $k$.
Update
So, here our sum is:
$$\displaystyle \sum_{k=0}^\infty A^k = \sum_{k=0}^\infty \begin{bmatrix}-\frac{1}{2} & -2 & -1\\0 & \frac{1}{3} & 0\\ 0 & 0 & -\frac{1}{2}\end{bmatrix}^k = \sum_{k=0}^\infty  \begin{bmatrix}(-1/2)^k & \left(\frac{4}{5}\right) 3^{1-k} ((-3/2)^k-1) & (-1)^k 2^{1-k} k \\ 0 & 3^{-k} & 0 \\ 0 & 0 & (-1/2)^k)\end{bmatrix}$$
Now, how do you evaluate each of these sums and do they each converge?
Update 2
You should get:
$$\displaystyle \sum_{k=0}^\infty A^k = \begin{bmatrix}\frac{2}{3} & -2 & -\frac{4}{9}\\0 & \frac{3}{2} & 0\\0 & 0 & \frac{2}{3}\end{bmatrix}$$
I would also recommend taking this answer and seeing how the other answers apply.
A: Choose any algebra norm $||.||$ and if $||A||<1$ then the geometric series 
$$\sum_{k=0}^\infty ||A||^k$$
is convergent so the series $\displaystyle\sum_{k=0}^\infty A^k$ is normal convergent and then it's convergent.
One possible method to calulate the limit: let $\pi_A$ the minimal polynomial of $A$ and by the Euclidean division we have
$$x^k=Q(x)\pi_A(x)+R_k(x)$$
with $\deg(R_k)<\deg(\pi_A)$
and we have $$\sum_{k=0}^\infty A^k=\sum_{k=0}^\infty R_k(A)$$
and the last sum is more simple to calculate.
Added
In our case, the matrix $A$ is triangular and since $(A+\frac{1}{2}I_3)(A-\frac{1}{3}I_3)\neq0$ so we can see easly that the minimal polynomial is
$$\pi_A(x)=\frac{1}{12}(3x-1)(2x+1)^2$$
hence in the Euclidean division the remainder polynomial is
$$R_k(x)=a_k x^2+b_k x+c_k$$
and we have
$$\begin{align}\\
\left(\frac{1}{3}\right)^k&=a_k/9+b_k/3+c_k\\
\left(\frac{1}{2}\right)^k&=a_k/4-b_k/2+c_k\\
k\left(\frac{-1}{2}\right)^{k-1}&=-a_k+b_k\\
\end{align}$$
so we solve the above system for $a_k,b_k$ and $c_k$ and finally we find
$$\sum_{k=0}^\infty A^k=\left(\sum_{k=0}^\infty a_k\right)A^2+\left(\sum_{k=0}^\infty b_k\right)A+\left(\sum_{k=0}^\infty c_k\right)I_3$$
A: A special solution to this problem:
You may write $A = D+N$, where $D$ is diagonal and $N$ is nilpotent in an obvious way. Then it is easy to see (either by a direct calculation or by the celebrated Cayley-Hamilton theorem) that $N^3 = O$ and thus we can simply $A^k$ as
$$ A^k = D^k + kD^{k-1}N + \frac{k(k-1)}{2}D^{k-2}N^2. $$
Now both the convergence and the summation value can be easily analyzed from the knowledge on power series in $\Bbb{R}$.
