You want first to fix any typographical errors (such as the unbalanced parentheses) and it's also wise to avoid symbol names beginning with capital letters. Then, to obtain a series expansion in powers of $1/z$, expand the expression around infinity, not zero:
Series[a + b (1 - Exp[-t/(b c)]/(z - Exp[-t/(b c)])) , {z, Infinity, 5}]
$(a+b)-\frac{b e^{-\frac{t}{b c}}}{z}-\frac{b e^{-\frac{2 t}{b c}}}{z^2}-\frac{b e^{-\frac{3 t}{b c}}}{z^3}-\frac{b e^{-\frac{4 t}{b c}}}{z^4}-\frac{b e^{-\frac{5 t}{b c}}}{z^5}+O\left[\frac{1}{z}\right]^6$
To confirm this, we could also replace $z$ by $1/z$, expand the series in non-negative powers of $z$, and then substitute $1/z$ back for $z$ once more:
Series[a + b (1 - Exp[-t/(b c)]/(z - Exp[-t/(b c)])) /. z -> (1/z), {z, 0, 5}] /. z -> 1/z
$(a+b)-\frac{b e^{-\frac{t}{b c}}}{z}-b e^{-\frac{2 t}{b c}} \left(\frac{1}{z}\right)^2-b e^{-\frac{3 t}{b c}} \left(\frac{1}{z}\right)^3-b e^{-\frac{4 t}{b c}} \left(\frac{1}{z}\right)^4-b e^{-\frac{5 t}{b c}} \left(\frac{1}{z}\right)^5+O\left[\frac{1}{z}\right]^6$
The two results are clearly equivalent expressions of the same series. If the terminal O[]
term is undesirable, remove it by applying Normal
to the output.
Series[a + b (1 - k/(z - k)), {z, 0, 5}]
gives $(a+2 b)+\frac{b z}{k}+\frac{b z^2}{k^2}+\frac{b z^3}{k^3}+\frac{b z^4}{k^4}+\frac{b z^5}{k^5}+O\left(z^6\right)$ $\endgroup$