How to calculate $\lim_{n\to\infty}\frac{p_n}{q_n}$ where $p_{n+1}=p_{n}+2q_{n},q_{n+1}=p_{n}+q_{n},p_{1}=p_{1}=1$ I tried to prove that $|a_{n+1}-a_{n}|<r|a_{n}-a_{n-1}|$ where $r$ is some number less than $1$. But I can't manage it. Would you help me slove this problem. Best regards!
 A: Hint. Let $x_n=p_n/q_n$ then $x_1=1$ and for $n\geq 1$,
$$x_{n+1}=f(x_n)=\frac{x+2}{x+1}.$$
Hence if $(x_n)_n $ is convergent to $L\in \mathbb{R}$ then $L=f(L)$.
Moreover note that 
$$f(x)-\sqrt{2}=\frac{(\sqrt{2}-1)}{x+1}\cdot (x-\sqrt{2}).$$
Can you take it from here?
A: Assuming the limit exists, say $\lim_{n\to\infty}\frac{p_n}{q_n}=L$, then consider the following:
$$L=\lim_{n\to\infty}\frac{p_{n+1}}{q_{n+1}}=\lim_{n\to\infty}\frac{p_n+2q_n}{p_n+q_n}=\lim_{n\to\infty}1+\frac{q_n}{p_n+q_n}=\lim_{n\to\infty}1+\frac{1}{1+p_n/q_n}=1+\frac{1}{1+L},$$
So $$L=1+\frac{1}{1+L},$$
which gives $L=\pm\sqrt{2}$, but we can easily rule out $L=-\sqrt{2}$.
I must stress that we are assuming that the limit exists...
A: use that $$p_n=\frac{1}{2}\left((1-\sqrt{2})^n+(1+\sqrt{2})^n\right)$$ and
$$q_n=-\frac{(1-\sqrt{2})^n-(1+\sqrt{2})^n}{2\sqrt{2}}$$
A: Clearly $\lim_{n\to\infty}p_n=\lim_{n\to\infty}q_n=\infty$. Let $x=\lim_{n\to\infty}\frac{p_n}{q_n}$. Thus by the Stolz-Cesaro theorem,
\begin{eqnarray}
x=\lim_{n\to\infty}\frac{p_n}{q_n}=\lim_{n\to\infty}\frac{p_{n+1}-p_n}{q_{n+1}-q_n}=\lim_{n\to\infty}\frac{2q_n}{p_n}=\frac{2}{x}
\end{eqnarray}
and hence $x=\sqrt 2$ since $p_n>0,q_n>0$.
