For what $\beta$ does the geometric series $I + A + A^2 + A^3 + \cdots$ converge to $(I-A)^{-1}$ 
Problem. Consider the matrix $A$ below.
$$A = \begin{bmatrix} {\beta\over 2} & 0\\ {\beta\over 2} & \beta
 \end{bmatrix}$$
For what $\beta$ does the sequence $I + A + A^n + \cdots$ converge to
$(I-A)^{-1}$? How quickly does the sequence converge, as a function of
$\beta$?


As we're only in the second week of an undergraduate course and haven't yet covered them, I'm not able to make use of matrix norms (in which case clearly it must be that $lim_{n\to\infty} |A_n| = lim_{n\to\infty} A_n = 0$, yielding $0<\beta<2$ by examination of the general form of $A^n$), and I'm not quite sure how else to approach this. Here's what I have so far.
Notice first that the general form of $A^n$ seems to be given by:
$$A^n = \begin{bmatrix}{\beta^n\over 2^n} &0\\ {\beta^n \over 2^n}(2^n-1)& \beta^n\end{bmatrix}$$
Notice also that $(I-A^n)^{-1}$ is then given by:
$$\Bigg(\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} - \begin{bmatrix}{\beta^n\over 2^n} &0\\ {\beta^n \over 2^n}(2^n-1)& \beta^n\end{bmatrix}\Bigg)^{-1} = \begin{bmatrix}1 + {\beta^n\over 2^n} & 0 \\ ({\beta\over 2})^n(2^n-1) & \beta^n+1\end{bmatrix}^{-1} = \begin{bmatrix} {2^n\over {2^\beta +\beta^n}} & 0\\ {{\beta^n(1-2^n)}\over (2^n\beta^n)(\beta^n+1)} & {1\over \beta^n +1}\end{bmatrix}$$
Now, if we look at the sum $A + A^2 + A^3 + \ldots$, it looks like:
$$I+A = \begin{bmatrix}
{\beta+2\over 2} & 0\\
{\beta\over 2} & \beta+1
\end{bmatrix}\;\;, \;\; I+A+A^2 = 
\begin{bmatrix}
{\beta^2 + 2\beta +4\over 4} & 0\\
{\beta(3\beta+2)\over 4} & \beta^2+\beta+1\\
\end{bmatrix}$$
It isn't clear to me that this series is approaching $(I-A^n)^{-1}$, nor how one would determine for what $\beta$ this holds. Moreover, it isn't clear what is meant by "how quickly" the series converges, nor how one would determine that as a function of $\beta$.
Where do I go from here?
 A: A sequence of matrices converge if and only if all of its coefficients are convergent series. So I think that you are almost done and that you can even compute the limit. Since you found that
$$A^n = \begin{bmatrix}{\beta^n\over 2^n} &0\\ {\beta^n \over 2^n}(2^n-1)& \beta^n\end{bmatrix}$$
then you can sum
$$\sum_{k = 0}^n A^k = \begin{bmatrix}\sum_{k=0}^n ({\beta\over 2})^n &0\\ \sum_{k=0}^n {\beta^n} -\sum_{k=0}^n ({\beta\over 2})^n & \sum_{k=0}^n\beta^n\end{bmatrix}$$
The coefficients are geometric series whose convergence and limits can be determined explicitely. You can then check that the limit is indeed the inverse matrix of $(I-A)$. Note that all coefficients have to converge in order to get convergence of the matrix.
Another way to deal with this question is to use the fact that a geometric series has an explicit remainder. Note that
$$
(I-A)(\sum_{k=0}^n A^k) = I - A^{n+1}
$$
From there, you can study the convergence of $A^{n+1}$ with the help of the expression you have found, and determine for which $\beta$ this goes to the zero matrix.
