real analysis - Evaluate $lim_{nrightarrowinfty} frac{a_n}{b_n}$

Let $a_n$ and $b_n$ be a recursive sequence with seed value $a_0=0,a_1=1$, $b_0=1$ and $b_1=2$ such that

$$\begin{align} \\ &a_{n+1}=(4n+2)a_n+a_{n-1}\\\\&b_{n+1}=(4n+2)b_n + b_{n-1} \end{align}$$

Find $\displaystyle\lim_{n\rightarrow\infty} \frac{a_n}{b_n}$. (Ans. $\frac{e-1}{e+1}$)

I don't know how to start. Any help would be appreciated.

Answer

Hint: the continued fraction of $\tanh\left(\frac{1}{2}\right)$ is given by:
$$\tanh\left(\frac{1}{2}\right)=[0;2,6,10,14,18,22, 26, 30, 34, 38, 42,\ldots]\tag{1}$$
due to Gauss' continued fraction, and your sequence $\left\{\frac{a_n}{b_n}\right\}_{n\geq 1}$ is just the sequence of convergents of the RHS of $(1)$.

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If you change the initial values $a_0,a_1,b_0,b_1$, the limit takes the form $\frac{a+bz}{c+dz}$ with $z=\tanh\left(\frac{1}{2}\right)$ by the general theory of continued fractions.