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.

  • $\begingroup$ 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? $\endgroup$ – Fimpellizieri Feb 12 '17 at 16:37
  • $\begingroup$ I don't know, sir. $\endgroup$ – MoNtiDeaD MoonDogs Feb 12 '17 at 16:40
  • $\begingroup$ 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 ? $\endgroup$ – Donald Splutterwit Feb 12 '17 at 17:00
  • $\begingroup$ the answer is $\frac{e-1}{e+1}$ but i don't know how to solve. $\endgroup$ – MoNtiDeaD MoonDogs Feb 12 '17 at 17:05
  • $\begingroup$ 0.46211716 by calculation ... in agreement with the value you state ... this is very similar to Euler's continued fraction for e. $\endgroup$ – 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.

  • $\begingroup$ I'm very confused how I get $\tanh \frac{1}{2}$ that is the limit. $\endgroup$ – MoNtiDeaD MoonDogs Feb 12 '17 at 17:38
  • $\begingroup$ @MoNtiDeaDMoonDogs: I just recognized by experience a well-known continued fraction. $\endgroup$ – Jack D'Aurizio Feb 12 '17 at 17:39
  • $\begingroup$ @MoNtiDeaDMoonDogs: see also here: topologicalmusings.wordpress.com/2008/08/04/… $\endgroup$ – Jack D'Aurizio Feb 12 '17 at 17:40
  • $\begingroup$ If I change the seed value, what should i get the limit. i mean that different seed value, it may imply different value of limit. $\endgroup$ – MoNtiDeaD MoonDogs Feb 12 '17 at 17:52
  • $\begingroup$ @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. $\endgroup$ – Jack D'Aurizio Feb 12 '17 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.