Exponential of a non terminating matric So I understand how to calculate the exponential of matrices that eventually terminate; however, how to approach the cases in which the matrix does not seem to truncate? For example with the matrix $M=$$\quad
\begin{pmatrix} 
2 & 1 \\
0 & 2 
\end{pmatrix}$
I have calculated the first few terms: for $k=1$ $\quad
\begin{pmatrix} 
2 & 2 \\
0 & 2 
\end{pmatrix}$
for $k=2$ $\quad
\begin{pmatrix} 
4/3 & 2 \\
0 & 4/3 
\end{pmatrix}$
for $k=5$ $\quad
\begin{pmatrix} 
4/15 & 2/3 \\
0 & 4/15 
\end{pmatrix}$
Any advice on summing these together or developing a formula that would help me with the summation?
Similarly for a 3x3 such as $\quad
\begin{pmatrix} 
2 & 2 & 0\\
0 & 2 & 1\\
0 & 0 & 2
\end{pmatrix}$
I'm really not sure how to find $e^M$ for this.
 A: The basic observation that can be used to compute the exponent explicitly is that if $X,Y$ are matrices that commute ($XY = YX$) then $\exp(X + Y) = \exp(X) \exp(Y)$.
Your matrix $M$ can be written as $M = D + N$ where 
$$ D = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \,\,\, N= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. $$
Note that $D$ is diagonal and so the exponent of $D$ is easy to compute:
$$ \exp(D) = \begin{pmatrix} \exp(1) & 0 \\ 0 & \exp(1) \end{pmatrix} = \begin{pmatrix} e & 0 \\ 0 & e \end{pmatrix}. $$
The matrix $N$ is nilpotent satisfying $N^2 = 0$ and so the exponent of $N$ "eventually terminates":
$$ \exp(N) = I + N + \frac{N^2}{2} + \dots = I + N = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}. $$
Thus,
$$ \exp(M) = \exp(D) \exp(N) = \begin{pmatrix} e & 0 \\ 0 & e \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} e & e \\ 0 & e \end{pmatrix}. $$   
This also works for your second matrix (and more generally for any upper triangular matrix). For a general matrix, you won't be necessarily able to decompose $M = D + N$ where $D$ is diagonal and $N$ is nilpotent but if you work over the complex numbers, the Jordan decomposition tells you that $M$ is similar to a matrix of the form $D + N$ so you can find an invertible $P$ with $P^{-1}MP = D + N$ and use the property that $\exp(PAP^{-1}) = P\exp(A)P^{-1}$. More explicitly, $M = P(D+N)P^{-1}$ and
$$ \exp(M) = \exp(P(D+N)P^{-1}) = P\exp(D)\exp(N)P^{-1}. $$
