Where to find the proof of this property? 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:


*

*$\sum\limits_{n=1}^{+\infty} f_n = 1$.

*$\gcd(i : f_i > 0) = 1$.
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!
 A: In probabilistic terms, the result you want is precisely the so-called local renewal theorem, lattice version, 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? 
