If $A=C-\frac{1}{2}C^2+\frac{1}{3}C^3-\frac{1}{4}C^4+\cdots$ where
C=$\left( \begin{array}{ccc} 0 & c & c^2 \\ 0 & 0 & c \\ 0 & 0 & 0 \\ \end{array} \right)$
(i) I want to show that A possesses finitely many terms apart from $0$. (ii) How can I compute A? (iii) How can I also illustrate that $A-\frac{1}{2!}A^2+\frac{1}{3!}A^3-\frac{1}{4!}A^4+\cdots$ exhibits finitely many non-zero terms. (iv) How can I also show that $A-\frac{1}{2!}A^2+\frac{1}{3!}A^3-\frac{1}{4!}A^4+\cdots=C$?
On
A strictly upper triangular matrix like $C$ is always nilpotent, meaning that $C^k=0$ for some finite $k$. More generally, $M^n=0$ for any $n\times n$ strictly upper (or lower) triangular matrix, then $C^3=0$ and $$ A= C - \frac{1}{2}C^2. $$
This answers (i) and (ii). As $A$ is a polynomial of a nilpotent matrix, it is itself nilpotent, what answers (iii). Now realize that $A^3=0$ and $A^2=C^2$, then $$ A-\frac{1}{2}A^2=C-C^2\neq C. $$ Thus (iv) is not correct.
Hint:
You may suppose $c\ne0$. Rewrite $C$ as $\;c\begin{pmatrix}0&1&c\\0&0&1\\0&0&0\end{pmatrix}$, and check $C^3=0$, so $A$ is indeed a finite sum.