Let $p\_n$ be a probability distribution on the positive
integers $n$. Let
 $$ \frac{1}{1-\sum\_{n\geq 1} p\_nx^n}=\sum\_{k\geq 0}a\_kx^k. $$
Suppose there does not exist an integer $d>1$ such that
$d|n$ whenever $p\_n\neq 0$. I remember once seeing a proof
of the result
 $$ \lim\_{k\to\infty} a\_k = \frac{1}{\sum\_{n\geq 1}np\_n}, $$
but I cannot recall the proof. Can someone provide such a
proof? Problem A6 on the 1960 Putnam exam is the case
$p\_1=\cdots=p\_6=1/6$. The result is intuitively clear, since
if we pick integers $n\_1,n\_2,\dots$ from the distribution
$p$, then $a\_k$ is the probability that some $n\_1+n\_2+\cdots
+ n\_j=k$. Now $E:=\sum\_{n\geq 1}np\_n$ is the expected value of
$k$. Thus on the average we are picking every $E$th positive
integer, so a proportion $1/E$ should be chosen.


**Note.** One way to prove the formula would be to show that
the function
 $$ \frac{1}{1-\sum\_{n\geq 1} p\_nx^n} -\frac{1/E}{1-x} $$
has radius of convergence greater than 1.



A limit obtained from a probability distribution on the positive integers
This is proved in Spitzer's "Principles of Random Walk", claim P3 of Section 9 (p.100 of the second edition).


