A very curious rational fraction that converges. What is the value?

If $\mu=\lim_{n\to\infty} \frac{a_{n} }{b_{n} } $exists then $$ \mu=\lim_{n\to\infty} \frac{a_{n+1} }{b_{n+1} } = \lim_{n\to\infty} \frac{b_n+2a_n + 14}{9b_n+ 2a_n+70 }\\= \lim_{n\to\infty} \frac{\frac{b_{n} }{b_{n} }+2\frac{a_{n} }{b_{n} } + \frac{14 }{b_{n} }}{9\frac{b_{n} }{b_{n} }+ 2\frac{a_{n} }{b_{n} }+\frac{70 }{b_{n} } }\\ =\frac{1+2\lim_{n\to\infty}(\frac{a_{n} }{b_{n} }) + \lim_{n\to\infty}(\frac{14 }{b_{n} })}{9+ 2\lim_{n\to\infty}(\frac{a_{n} }{b_{n} })+\lim_{n\to\infty}(\frac{70 }{b_{n} }) }\\ =\frac{1+2\mu+ 0}{9+ 2\mu+0} $$ Here we used the fact that ${b}_{{n}}>{n}$ and therefore $$0\leq~\lim_{{n}\to\infty}~{\frac{1}{{b}_{{n}}}}<\lim_{{n}\to\infty}~{\frac{1}{{n}}}\leq0$$

So $$2\mu^2+7\mu-1=0$$. But $\mu \gt 0$, so $$\mu={{\sqrt{57}-7}\over{4}} $$

One idea to get the limit in closed form :

Let $A=\begin{pmatrix} 2 & 1 \\ 2 & 9 \end{pmatrix}$, $X_n = \begin{pmatrix} a_n \\ b_n \end{pmatrix}$, and $b=\begin{pmatrix} 14 \\ 10 \end{pmatrix}$.

You can write $X_{n+1} = AX_n + b$. The idea is to solve the equation $X=AX+b$ ($X$ being a two-dimensional vector) - this equation has a unique solution as $(A-I)$ is non-singular. Let's call $X$ this solution; you can write $(X_{n+1}-X)=A(X_n - X)$ then, for all $n$, you have :

\begin{equation} (X_{n} - X ) =A^n (X_{0} - X) \end{equation}

$A^n$ can be evaluated by diagonalizing $A$. This will give you $X_n$ (and then $a_n$ and $b_n$ in closed form).