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.