I am doing some exercises in the analytic and there is a problem as following: 


``Let $\{f\_n\}\_{n \in \mathbb{ N}}$'' to be a positive sequence such that:


1. $\sum\limits\_{n=1}^{+\infty} f\_n = 1$.
2. $\gcd(i : f\_i > 0) = 1$.


Suppose that $\mu = \sum\limits\_{n =1}^{+ \infty} nf\_n < + \infty$.
If the sequence $(u\_n)\_{n\in \mathbb{N}}$ is defined by: $$u\_0 := 1; \ u\_n := \sum\limits\_{k =1}^n f\_k u\_{n-k}; \ n =1, 2, 3, \ldots $$ then $\lim\limits\_{n \to + \infty} u\_n = \dfrac{1}{\mu} \ . $'' 


Could you tell me about the reference or a proof of this property? Thank you so much!



Where to find the proof of this property?
In probabilistic terms, the result you want is precisely the so-called [local renewal theorem, lattice version](https://projecteuclid.org/download/pdf_1/euclid.pjm/1103051394), due to Kolmogorov '36, Erdös--Feller--Pollard '49, and Chung--Wolfowitz '52. Indeed, let $S\_1,S\_2,\dots$ be independent random variables such that $P(S\_n=s)=f\_s$ for all natural $n,s$, with $\gcd\{s\colon f\_s>0\} = 1$. For $t=0,1,\dots$, let 
$$X\_t:=\sum\_{n=0}^\infty 1\_{J\_n=t},$$
where $J\_n:=\sum\_1^n S\_i$, with $J\_0:=0$. Then the local renewal theorem states that 
$$u\_t:=EX\_t=\sum\_{n=0}^\infty P(J\_n=t)\to\frac1{ES\_1}=\frac1\mu$$
as $t\to\infty$. 


On the other hand, conditioning on $J\_1[=S\_1]$, we see that the numbers 
$$u\_t=EX\_t$$
indeed satisfy your conditions,
\begin{equation}
 u\_0=1; \quad u\_t=\sum\_{s=1}^t f\_s u\_{t-s}; \quad t =1,2,\dots, \tag{1}
\end{equation}
which determine the $u\_t$'s uniquely. 


Thus, you have your desired result. 




---


**Details on (1):** First here, $X\_0=1\_{J\_0=0}=1$, whence $u\_0=EX\_0=1$. Next, for $t=1,2,\dots$
\begin{multline}
 u\_t=EX\_t=\sum\_{n=0}^\infty P(J\_n=t)=\sum\_{n=1}^\infty P(J\_n=t) \\ 
 =\sum\_{n=1}^\infty \sum\_{s=1}^t P(J\_1=s)P(J\_n-J\_1=t-s)
 =\sum\_{n=1}^\infty \sum\_{s=1}^t P(J\_1=s)P(J\_{n-1}=t-s) \\ 
 =\sum\_{s=1}^t P(J\_1=s)\sum\_{n=1}^\infty P(J\_{n-1}=t-s)
 =\sum\_{s=1}^t f\_s u\_{t-s}, 
\end{multline}
thus completing the confirmation of (1). 




---


**Comment:** One can also try to obtain a purely analytic proof, using generating functions and the Cauchy integral formula, as follows. By (1) and induction, $0\le u\_t\le1$ for all $t$. Also, clearly $0\le f\_s\le1$ for all $s$. So, we may consider 
\begin{equation}
 f(z):=\sum\_{n=1}^\infty f\_n z^n,\quad u(z):=\sum\_{n=0}^\infty u\_n z^n
\end{equation}
for $z$ with $|z|<1$. For such $z$, by (1),
\begin{multline}
 u(z)=\sum\_{n=0}^\infty u\_n z^n=1+\sum\_{n=1}^\infty z^n\sum\_{k=1}^n f\_k u\_{n-k}
 =1+\sum\_{n=1}^\infty \sum\_{k=1}^n f\_k z^k u\_{n-k} z^{n-k} \\ 
 =1+\sum\_{k=1}^\infty f\_k z^k \sum\_{n=k}^\infty u\_{n-k} z^{n-k}
 =1+f(z)u(z),
\end{multline}
whence 
\begin{equation}
 u(z)=\frac1{1-f(z)}. 
\end{equation}
So, by the Cauchy integral formula, 
\begin{equation}
 u\_n=\frac1{2\pi i}\,\oint\_{C\_r}\frac{dz}{(1-f(z))z^{n+1}}
 =\frac1{2\pi r^n}\,\int\_0^{2\pi}\frac{e^{-in\theta}d\theta}{1-f(re^{i\theta})},
\end{equation}
where $r\in(0,1)$ and $C\_r$ is the circle of radius $r$ centered at $0$. 
Heuristically, since $1-f(z)\sim\mu(1-z)$ as $z\to1$ while $|z|\le1$, one may now expect that 
\begin{equation}
 u\_n\approx\frac1{2\pi i}\,\oint\_{C\_r}\frac{dz}{\mu(1-z)z^{n+1}}=\frac1\mu. 
\end{equation}
However, at this point I feel not experienced enough in such analyses to show rigorously this way that $u\_n\to\frac1\mu$; in particular, it is unclear how to incorporate the gcd condition into such an analysis. Anyone? 


