A Proof of Basic Limit Theorem of Renewal Theory

Let ₙ\ₙ₌₀^∞⊂ [0,1] satisfy q₀=0, ∑ₙ₌₀^∞ qₙ=1, and ≥ 1 qₙ≠ 0\=1. We consider the following process: Let x be a real number. We first set $x=0$. Then x is increased by i with probability qᵢ~(i=0,1,2,⋯) every time. For n≥ 0, let pₙ be the probability such that $x=n$ occurs, so we have p₀=1 and pₙ=q₁pₙ₋₁+q₂pₙ₋₂+⋯+qₙp₀~(n≥ 1). In this setting, we have limₙ pₙ=1/∑ᵢ₌₀^∞ iqᵢ, where we define 1/∑ᵢ₌₀^∞ iqᵢ=0 if ∑ᵢ₌₀^∞ iqᵢ=+∞. This result is known as (discrete case of) Blackwell's renewal theorem. The proof of limₙ pₙ=1/∑ᵢ₌₀^∞ iqᵢ is not trivial, while the meaning of limₙ pₙ=1/∑ᵢ₌₀^∞ iqᵢ is clear since the expected value of increasing number i is ∑ᵢ₌₀^∞ iqᵢ. Many proofs of this result have been given. In this paper, we will also provide a proof of this result. The idea of our proof is based on Fourier-analytic methods and Tauberian theorems for almost convergent sequences, while we actually need only elementary analysis.