A Proof of Basic Limit Theorem of Renewal Theory

Let $\{q_n{\}}_{n=0}^{\mathrm{\infty }}\subset [0,1]$ satisfy $q_0=0$, ${\sum }_{n=0}^{\mathrm{\infty }}{q}_n=1$, and $\gcd \{n\ge 1\mid q_n\mathrm{\ne }0\}=1$. 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 $n\ge 0$, let $p_n$ be the probability such that $x=n$ occurs, so we have $p_0=1$ and $p_n=q_1{p}_{n-1}+q_2{p}_{n-2}+\cdots +q_n{p}_0(n\ge 1)$. In this setting, we have ${\lim }_n{p}_n=1\mathrm{/}{\sum }_{i=0}^{\mathrm{\infty }}i{q}_i$, where we define $1\mathrm{/}{\sum }_{i=0}^{\mathrm{\infty }}i{q}_i=0$ if ${\sum }_{i=0}^{\mathrm{\infty }}i{q}_i=+\mathrm{\infty }$. This result is known as (discrete case of) Blackwell’s renewal theorem. The proof of ${\lim }_n{p}_n=1\mathrm{/}{\sum }_{i=0}^{\mathrm{\infty }}i{q}_i$ is not trivial, while the meaning of ${\lim }_n{p}_n=1\mathrm{/}{\sum }_{i=0}^{\mathrm{\infty }}i{q}_i$ is clear since the expected value of increasing number $i$ is ${\sum }_{i=0}^{\mathrm{\infty }}i{q}_i$. 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.