A limit of a recursion involving the prime number function Suppose $A_1=1$, $B_1=2$ and  
$\left\{
\begin{array}{l}
A_{k+1}=A_k\cdot (p_{k+1}-1)+B_k\\
B_{k+1}=B_k\cdot p_{k+1}
\end{array}
\right.
$
where $p_k$ is the $k$-th prime number. I would like a formal proof of that
$\displaystyle\lim_{n\to\infty}\frac{A_n}{B_n}=1$.
I've checked it computationally, but am unsure about the proof.

Let $r_k=\frac{A_k}{B_k}$. We have $B_k=p_1\cdots p_k$ so
$$r_{k+1}-1=(r_k-1)\cdot\left(1-\frac1{p_{k+1}}\right).$$
It follows that $$r_n=1+\left(1-\frac1{p_1}\right)\cdots\left(1-\frac1{p_n}\right)$$
and it suffices to note that $\prod_p\left(1-\frac1p\right)=0$.

In general, when dealing with a linear recursion (with possibly non-constant coefficients) it is worth homogenizing it, that is, getting rid of the constant term. In this case however I discovered it in the following (deeper) way:
From $$r_{k+1}=r_k\cdot\left(1-\frac1{p_{k+1}}\right)+\frac1{p_{k+1}}$$
one may note that
$$r_k=1+\sum_{\substack{p\mid n\implies p\leq p_k\\ n\text{ even}}}\frac{\mu(n)}n.$$
Because $\displaystyle\sum_{\substack{p\mid n\implies p\leq p_k\\ n\text{ odd}}}\!\!\!\tfrac{\mu(n)}n=-2\sum_{\substack{p\mid n\implies p\leq p_k\\ n\text{ even}}}\!\!\!\tfrac{\mu(n)}n$ one sees that
$$r_k-1=\sum_{p\mid n\implies p\leq p_k}\frac{\mu(n)}n=\left(1-\frac1{p_1}\right)\cdots\left(1-\frac1{p_k}\right).$$
