# Evaluate $\lim_{n\rightarrow\infty} \frac{a_n}{b_n}$

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.

• You can try to solve the recurrence for each sequence (ie, find a formula), then plug into the limit. Do you know how to solve this kind of recurrence? – Fimpellizieri Feb 12 '17 at 16:37
• I don't know, sir. – MoNtiDeaD MoonDogs Feb 12 '17 at 16:40
• I calculate $0.45543809977901$ as the ratio. Not sure how to solve the recurrence relation ... someone give me a clue ? We could use generating functions ? – Donald Splutterwit Feb 12 '17 at 17:00
• the answer is $\frac{e-1}{e+1}$ but i don't know how to solve. – MoNtiDeaD MoonDogs Feb 12 '17 at 17:05
• 0.46211716 by calculation ... in agreement with the value you state ... this is very similar to Euler's continued fraction for e. – Donald Splutterwit Feb 12 '17 at 17:19

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 $(1)$.
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.
• I'm very confused how I get $\tanh \frac{1}{2}$ that is the limit. – MoNtiDeaD MoonDogs Feb 12 '17 at 17:38
• @MoNtiDeaDMoonDogs: if you change the initial values, you get something of the form $\frac{a+bz}{c+dz}$ where $z=\tanh\frac{1}{2}$. This follows from the general theory of continued fractions. – Jack D'Aurizio Feb 12 '17 at 17:55