4
$\begingroup$

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.

  • 0
    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? – 2017-02-12
  • 0
    I don't know, sir. – 2017-02-12
  • 0
    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 ? – 2017-02-12
  • 0
    the answer is $\frac{e-1}{e+1}$ but i don't know how to solve. – 2017-02-12
  • 0
    0.46211716 by calculation ... in agreement with the value you state ... this is very similar to Euler's continued fraction for e. – 2017-02-12
  • 0
    It's the answer, but I don't know how I get it. – 2017-02-12
  • 0
    another way to solve this is via generating functions... – 2017-02-13

1 Answers 1

5

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.

  • 0
    I'm very confused how I get $\tanh \frac{1}{2}$ that is the limit. – 2017-02-12
  • 0
    @MoNtiDeaDMoonDogs: I just recognized by experience a well-known continued fraction. – 2017-02-12
  • 0
    @MoNtiDeaDMoonDogs: see also here: https://topologicalmusings.wordpress.com/2008/08/04/continued-fraction-for-e/ – 2017-02-12
  • 0
    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. – 2017-02-12
  • 0
    @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. – 2017-02-12
  • 0
    I see but I'm still confused that I get the limit is $z=\tanh\frac{1}{2}$, but It's not $\frac{a+bz}{c+dz}$. I appreciate all your help and It's better if you show the value is $\tanh\frac{1}{2}$. – 2017-02-12
  • 0
    @MoNtiDeaDMoonDogs: you asked me a different thing, you asked me *what happens if we change the initial values?*. If we leave the initial values as they are, the sequence of $\frac{a_n}{b_n}$ is the sequence of convergents of a continued fraction, the RHS of $(1)$. But such continued fraction is well-known, hence the limit is given by the whole continued fraction, i.e. $\tanh\frac{1}{2}$. Is is clear now? – 2017-02-12
  • 0
    I'm sorry, sir. I know it's convergent and equal to $\tanh\frac{1}{2}$ that is continued fraction. if I change the initial values of $a_n$ or $b_n$, the limit will change to $\frac{a+bz}{c+dz}$, $z=\tanh\frac{1}{2}$. I would like to know that $a_0=0,a_1=1,b_0=1$ and $b_1=2$, I will get the value is $\tanh\frac{1}{2}$, but not $\frac{a+bz}{c+dz}$. – 2017-02-12
  • 0
    Is it the general initial value of sequence for limit or continued fraction? – 2017-02-12
  • 0
    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53519/discussion-between-montidead-moondogs-and-jack-daurizio). – 2017-02-12
  • 0
    @MoNtiDeaDMoonDogs: as already said, if you leave the initial values as they are defined in your question, you exactly get the sequence of convergents of the RHS of $(1)$. – 2017-02-12
  • 0
    Thank you, sir. I see the Generalized Continued Fraction. – 2017-02-12
  • 0
    I'm sorry for my confusion. – 2017-02-12