# Infinite series of matrices almost but not quite matrix exponential

I'm working on a problem that has brought up for me the need to address infinite series of the following form, $$\sum_{i=k}^\infty \frac{1}{i!}A^{i-k+1}$$ where $A$ is an $n\times n$ matrix. If $k = 0$, then this is clearly the power series of a matrix exponential. If $A$ were a real number, $x$, then it's easy to see that $$\sum_{i=k}^\infty \frac{1}{i!}x^{i-k+1} = \frac{1}{x^{k-1}}\left(e^x-\sum_{i=0}^k\frac{1}{i!}x^i \right),$$ by simply equating to the power series of the exponential function (at least I think I got those indices right).

I assume that the same thing would hold if $A$ were invertible; unfortunately, in my case $A$ is not invertible. However I have verified that my sums do converge to something, so I figured I'd ask if anyone had any idea of a way to simplify my infinite series.

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What do you mean by "simplify"? The right hand side of your scalar equation defines an entire function on the complex plane, which is no worse than $e^z$ in any respect except it lacks the multiplication identity... –  fedja Nov 9 '12 at 2:32
Exactly---I want something like the right hand side of my scalar equation, but for the matrix equation. Basically I want to reduce my matrix equation to something that isn't an infinite sum. –  Josh Nov 9 '12 at 2:44
Erm... But even $e^z$ is an infinite sum! $e^A$ is just a fancy name for it and it is neither better, nor worse than $F(A)$ where $F$ is the function defined by your equation... –  fedja Nov 9 '12 at 2:47
Okay, for context I am trying to approximate exp(Q) where I can decompose Q = A+B, and I can easily analytically compute exp(A) and exp(B). I want to make an approximation in which I am basically neglecting terms that are higher order than linear in B, and after doing a bunch of manipulations I end up with a bunch of terms that look like the series in question multiplied by B times some constant. So I'm hoping that if I can simplify these series, I will be able to come up with a slick approximation to exp(Q). –  Josh Nov 9 '12 at 15:12