A Proof of Basic Limit Theorem of Renewal Theory

Let \(\{q_n\}_{n=0}^\infty\subset [0,1]\) satisfy \(q_0=0\), \(\sum_{n=0}^\infty q_n=1\), and \(\gcd\{n\geq 1\mid q_n\neq 0\}=1\). We consider the following process: Let \(x\) be a real number. We first set \(x=0\). Then \(x\) is increased by \(i\) with probability \(q_i~(i=0,1,2,\cdots)\) every time. For \(n\geq 0\), let \(p_n\) be the probability such that \(x=n\) occurs, so we have \(p_0=1\) and \(p_n=q_1p_{n-1}+q_2p_{n-2}+\cdots+q_np_0~(n\geq 1)\). In this setting, we have \(\lim_n p_n=1/\sum_{i=0}^\infty iq_i\), where we define \(1/\sum_{i=0}^\infty iq_i=0\) if \(\sum_{i=0}^\infty iq_i=+\infty\). This result is known as (discrete case of) Blackwell's renewal theorem. The proof of \(\lim_n p_n=1/\sum_{i=0}^\infty iq_i\) is not trivial, while the meaning of \(\lim_n p_n=1/\sum_{i=0}^\infty iq_i\) is clear since the expected value of increasing number \(i\) is \(\sum_{i=0}^\infty iq_i\). 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.