Solve matrix exponential power series The matrix exponential is defined as
$$ e^A  = \sum_{n=0}^\infty \frac{1}{n!} A^n  $$
However I would like to solve something similar:
$$ B = \sum_{n=1}^\infty \frac{1}{n!} A^{n-1}  $$
(NOTE: starting index and the power)
I can transform that into
$$ B = A^{-1}\sum_{n=0}^\infty \frac{1}{n!} A^{n} -I = A^{-1} \left(e^A - I \right) $$



*

*That works. But when $A^{-1}$ does not exist ($A$ is not invertible) it does not work. 

*Is it possible to make it work for general $A$? Has anyone thoughts on that? Thanks.

Basically, your question asks how to apply the function $(e^x-1)/x$ to a matrix. One way to do it is using the integral representation
$$\frac{e^x-1}{x}=\int_0^1 e^{tx}\ dt.$$
Thus we can define $B$ to be
$$
B:=\int_0^1 e^{tA}\ dt.
$$
Since you already know what the matrix exponential is, this expression makes sense: the integral of a matrix is computed by taking the integral of each of its entries.
Note also that this coincides with the power series definition you have written, since $$\int_0^1 \frac{(tA)^{n}}{n!}\ dt=\frac{A^{n}}{(n+1)!}.$$
