Series sum using Z-transform

The Z-transform of the sequence x(k) = e^(ak) is

X(z) = Σ_{k=0}^∞ e^(ak) z^(−k) = Σ (e^a/z)ᵏ = z/(z − e^a),   |z| > e^a

The required series is X(z) evaluated at z = 1:

Σ e^(ak) = X(1) = 1/(1 − e^a)

Remark. This is a geometric series with ratio r = e^a. For a > 0 we have r > 1, so the series actually diverges. The value 1/(1 − e^a) = −1/(e^a − 1) is only a formal result (z = 1 lies outside the region |z| > e^a). The series converges, with sum 1/(1 − e^a), when a < 0 (and for the usual form e^(−ak) with a > 0 the sum is 1/(1 − e^(−a)) = e^a/(e^a − 1)).

Answer: formally 1/(1 − e^a); convergent only if a < 0.