Confusion regarding summing a matrix series I have got following series while working out for an iteration problem.
$X_{k+1} = (1 + x + x^{2}+\cdots x^{2^{k+2}-3})$, $k = 0, 1, 2\cdots $ and $\|x\|<1$. My question is could I write $X_{k+1} = (1- x)^{-1}$, when $k \to \infty$. 
Thanks for the help and time.
 A: Since $\|X\|<1$, you have $I+X+X^2+\cdots+X^n = (I-X)^{-1}(I-X^{n+1})$ (this is easy to verify noting that $I-X$ is invertible since all eigenvalues $\lambda$  of $X$ satisfy $|\lambda|<1$, and by multiplying both sides by $I-X$).
Since $\|X^n\| \le \|X\|^n$, it follows that $\lim_n X^n = 0$, from which it follows that $\lim_n \sum_{k=0}^{2^{k+2}-3} X^k =  \lim_n \sum_{k=0}^n X^k = (I-X)^{-1}$.
A: Given a matrix $X$, define the matrix $X_k$ by
$$
X_k=I + \sum^k_{j=1} X^j
$$
where $I$ is the identity matrix.  Then if you multiply that equation by $I-X$ it proves that
$$
\begin{align}
X_k\left(I-X\right) & = \left(I + \sum^k_{j=1} X^j \right)-\left(X + \sum^k_{j=1} X^{j+1} \right)\\
&=I-X^{k+1}
\end{align}
$$
Hence
$$
X_k = \left(I - X^{k+1}\right)\left(I - X\right)^{-1}
$$
so if $\text{lim}_{k \to \infty}X^k = 0$ for all elements of the matrix then 
$$
\text{lim}_{k \to \infty}X_k = \left(I - X\right)^{-1}
$$
So you can write $X_k = \left(I - X\right)^{-1}$ as $k \to \infty$ as long as all elements in the matrix $X^k$ tend to zero for large $k$.
