Help: infinite sum for matrices Suppose $G$ is an $n\times n$ matrix.
Can someone show me how you can carry
$$I + G x^{-1} + G^2 x^{-2} + G^3 x^{-3} + G ^4 x^{-4} + \cdots$$
to 
$$(xI - G)^{-1}x$$
without having to "divide" over another matrix thereby performing an illegal operation?
 A: First of all, note that, for all $x \ne 0$,
$xI - G = x(I - x^{-1}G), \tag{1}$
whence
$(xI - G)^{-1}x = (x(I - x^{-1}G))^{-1}x = (I - x^{-1}G)^{-1}x^{-1}x = (I - x^{-1}G)^{-1}. \tag{2}$
To make further progress, and connect this equation with the power series
$I + x^{-1}G + x^{-2}G^2 + x^{-3}G^3 + x^{-4}G^4 + \ldots \tag{3}$
we need some hypothesis which ensures (3) is a meaningful expression, i.e., the series converges.  This need may be met if we assume $\vert x \vert > \Vert G \Vert$, for then $\Vert x^{-1}G \Vert = \vert x^{-1} \vert \Vert G \Vert  = \vert x \vert^{-1} \Vert G \Vert < 1$; thus, in the usual manner, we write
$(I - x^{-1}G)\sum_0^m (x^{-1}G)^i = I - (x^{-1}G)^{m + 1}; \tag{4}$
with $\Vert x^{-1} G \Vert < 1$, we see that
$\Vert (x^{-1}G)^{m + 1} \Vert \le \Vert x^{-1} G \Vert^{m + 1} \to 0 \tag{5}$
as $m \to \infty$; this in turn implies the series (3) converges, since it is majorized in norm by the geometric series $\sum_0^\infty \Vert x^{-1}G \Vert^i$, and that
(4) in the limit becomes
$(I - x^{-1}G)\sum_0^ \infty(x^{-1}G)^i = I. \tag{6}$
(6) shows that
$(I - x^{-1}G)^{-1} = \sum_0^ \infty(x^{-1}G)^i, \tag{7}$
and combining this with (2) yields
$(xI - G)^{-1}x = \sum_0^ \infty(x^{-1}G)^i, \tag{8}$
as per request.
Note Added Saturday 14 February 2015 11:19 PM PST:  The clause "majorized in norm by the geometric series $\sum_0^\infty \Vert x^{-1}G \Vert^i$" which I used in the above, apparently, according to the comment of Math Newb (see below), could use further explication.  So here goes:  consider the sequence of partial sums
$S_m = \sum_0^m (x^{-1}G)^i; \tag{9}$
for $n > m$ we have
$S_n - S_m = \sum_{m + 1}^n  (x^{-1}G)^i, \tag{10}$
thus
$\Vert S_n - S_m \Vert = \Vert \sum_{m + 1}^n  (x^{-1}G)^i \Vert, \tag{11}$
and so, by the triangle inequality and $\Vert XY \Vert \le \Vert X \Vert \Vert Y \Vert$ for matrices $X$ and $Y$, 
$\Vert S_n - S_m \Vert \le \sum_{m + 1}^n  \Vert(x^{-1}G)^i \Vert \le  \sum_{m + 1}^n  \Vert(x^{-1}G) \Vert^i$
$= \Vert x^{-1}G \Vert^{m + 1} \sum_0^{n - m -1} \Vert x^{-1} G\Vert^i. \tag{12}$
The sum on the right is geometric, with ratio $\Vert x^{-1} G \Vert < 1$; it increases monotoically as $n - m - 1$ grows, being entirely comprised of positive terms.  Thus, as is well-known, we have
$\sum_0^{n - m -1} \Vert x^{-1} G\Vert^i < \sum_0^\infty \Vert x^{-1} G\Vert^i = \dfrac{1}{1 - \Vert x^{-1} G \Vert}, \tag{13}$
and it then follows that
$\Vert S_n - S_m \Vert < \dfrac{ \Vert x^{-1}G \Vert^{m + 1}}{1 - \Vert x^{-1} G \Vert} \tag{14}$
for any $m$ independently of $n > m$.  This shows that the sequence $S_m$ is Cauchy and hence converges, from which the rest of my argument follows.  This note is basically an expansive expansion of the definition of the phrase "majorized in norm"; and, hoping the bots and other censors sleep at this late hour, I add I truly hope this helps the OP decipher my words.  End of Note.
