Kendall's renewal theorem: geometric convergence of $v_n$ forces $\sum_k a_k\kappa^k<\infty$

Kendall's renewal theorem, the geometric-convergence direction. Let $a = (a_k)_{k \ge 1}$ be a probability distribution on the positive integers ($a_k \ge 0$, $a_0 = 0$, $\sum_k a_k = 1$), and let $v$ be its renewal sequence: $v_0 = 1$ and

so that $v_n$ is the probability that a renewal occurs at time $n$ when inter-renewal times are i.i.d. with law $a$. Suppose $v$ converges geometrically to a positive limit: there are $v_\infty > 0$, $M$ and $\rho \in [0,1)$ with

Then the inter-renewal law has a geometric tail: there is $\kappa > 1$ with

This is the implication (i) $\Rightarrow$ (ii) of Kendall's Theorem as stated in Meyn and Tweedie (Theorem 15.1.1), and it is the analytic engine behind the equivalence "geometric ergodicity $\Leftrightarrow$ geometric drift" (their Theorem 15.0.1) that the source invokes in Remark 1. In the present mission it is what the remaining open leaf geoDriftCondition_of_geometricallyErgodic needs: applied to the renewal sequence $v_k = \varepsilon\,(Q P^{n_0(k-1)})(C)$ of a small set $C$ with minorization $P^{n_0}(x,\cdot) \ge \varepsilon Q$ on $C$ (and $Q$ concentrated where the geometric rate constant is bounded), it turns the geometric total-variation convergence at the points of $C$ into geometric tails for the regeneration times, which is condition (ii) of Theorem 15.0.1 and yields the drift function $V(x) = E_x\bigl[\sum_{k \le \sigma_C} \kappa^k\bigr]$.

Proof idea. Write $A(z) = \sum_k a_k z^k$ and $V(z) = \sum_n v_n z^n$ for $|z| < 1$; the renewal equation is $V(z)(1 - A(z)) = 1$. The hypothesis says $H(z) = \sum_n (v_n - v_\infty) z^n$ is analytic on $|z| < 1/\rho$, so $G(z) := (1-z)V(z) = v_\infty + (1-z)H(z)$ extends analytically to $|z| < 1/\rho$ and satisfies $G(z)(1 - A(z)) = 1 - z$ on the open unit disc. $G$ has no zero on the closed unit disc: $G(1) = v_\infty > 0$, and at a point $z_0 \neq 1$ of the unit circle a zero of $G$ would force $|1 - A(r z_0)| = |1 - r z_0|/|G(r z_0)| \to \infty$ as $r \uparrow 1$, contradicting $|A| \le 1$. By compactness $G$ is zero-free on a disc $|z| < r$ with $r > 1$, so $1 - (1-z)/G(z)$ is analytic there and coincides with $A$ on the unit disc; uniqueness of power series coefficients and Cauchy's estimates give $\sum_k a_k \kappa^k < \infty$ for every $1 < \kappa < r$. No aperiodicity hypothesis is needed: it is implied by the convergence of $v_n$ to a positive limit.

Formalization Note The renewal recursion is written with the index shift $v_{n+1} = \sum_{j=0}^{n} a_{j+1} v_{n-j}$ to keep natural-number subtraction honest. The distribution is normalised by HasSum a 1; the limit $v_\infty$ must be positive (for a defective or periodic law the renewal sequence does not converge geometrically to a positive limit, so nothing is lost). The conclusion is stated as summability of $a_k \kappa^k$ for some $\kappa > 1$, the form in which it is consumed (a finite geometric moment of the inter-renewal time).