When $| \cal I| \sim 1000$, the method seems efficient. But what if $| \cal I| \ll 1000$ ? In this example, $| \cal I| = 3 \ll 1000$.

@PietroMajer, yeah, your series is simpler than the $cAMA'$. What I felt somewhat counter-intuitive is that why the simpler version of $cAMA'$ cannot play repeated squaring tricks. Its computational speedup seems slower than the original $cAMA'$.

@PietroMajer, for example, $X:=({(\cal I + L)} {(\cal I + L^2)}{(\cal I + L^4)} (...))I$, can we derive ${\cal L^4} I$ from ${\cal L^2} I$ efficiently just as repeated squaring does?

@PietroMajer, based on your ${\cal L}M$, it seems hard to utilized repeated squaring to calculate the series; whereas user35593's series (if $D$ is determined) can still play repeated squaring tricks. Is that right?

@PietroMajer, Thanks for pointing out the right convergence condition. Currently, what I am baffling with user35593's answer is that how to find out $D$ effectively. It seems to me that calculating $D$ is still very hard.

thanks. What if $m$ is large? Any efficient methods?

$A'$ denotes the transpose of matrix $A$