Prove $\sum\limits_{k=0}^{\infty} x^{k} = \frac{1}{1-x}$ also holds for matrices. I need to prove $\sum\limits_{k=0}^{\infty} x^{k} = \frac{1}{1-x}$ also holds for matrices. Thus the previous should be true when $x$ is a matrix. I honestly have no idea where to start, so any suggestions are welcome.
 A: I'll give hints, which are similar to what the comments suggest. 
To start, we need to suppose that $x$ is square, and $\|x\|<1.$ Call $S_n=I+x+x^2+\cdots+x^n.$ First, we claim that this converges as $n\rightarrow\infty,$ and the series converges absolutely and uniformly. Check why this is true. Call this limit $S$. We can see that $I-x$ and $S_n$ commute. What does their product give us, and what happens in the limit?
A: The statement you have written isn't necessarily true. In fact, you will need another condition that $\rho(A) < 1$ (spectral radius of $A$) to allow the series to converge. 
Here's the idea:
Let 
$$S_k = \sum_{n=0}^k A^n$$
Then 
$$S_k(I-A) = (I + A + \dots + A^k)(I-A) = I + A + \dots + A^k - (A + A^2 + \dots + A^{k+1}) = I - A^{k+1}$$
Similar one can see that $(I-A)S_k = I-A^{k+1}$. Now since $\rho(A) < 1$ then we have that $I-A$ is invertible and further that $\lim_{k \to \infty} A^k= 0$ thus taking the limit of both sides and define $S := \sum_{n=0}^{\infty} A^n = \lim_{k \to \infty} S_k$ we have that 
$$S(I-A) = I$$
and also 
$$(I-A)S = I$$
and therefore $S = (I-A)^{-1}$ which we wanted to show.  I believe that the converse also holds true. That is, 
$$\rho(A) < 1 \leftrightarrow \sum_{n=0}^{\infty} A^n \text{ converges and } \sum_{n=0}^{\infty} A^n = (I-A)^{-1}$$
