two recursive sequences and the limit of their quotient

Question

The sequences $\left \{ a_{n} \right \}$ and $\left \{ b_{n} \right \}$ are defined by the following recurrence relations:

1. $a_{1}=b_{1}=1$
2. $a_{n+1}=a_{n}+2b_{n}$
3. $b_{n+1}=a_{n}+b_{n}$

What is the limit of $\displaystyle \{\frac{a_{n}}{b_{n}}\}$ as n goes to infinity?

Answer 1

A good idea for this problem is to write $\frac{a_{n+1}}{b_{n+1}}$ in terms of $\frac{a_n}{b_n}$.

$\frac{a_{n+1}}{b_{n+1}}=\frac{a_n+2b_n}{a_n+b_n}=1+\frac{b_n}{a_n+b_n}=1+\frac{1}{\frac{a_n+b_n}{b_n}}=1+\frac{1}{1+\frac{a_n}{b_n}}$.

Now we write the sequence of numbers in the form $\frac{a_n}{b_n}$ as $c_n$. We get the recursion $c_{n+1}=1+\frac{1}{1+c_n}=\frac{2+c_n}{1+c_n}$

If a recursion converges to something, that means that it will eventually be really really close to that, which means that it basically does not change already (the changes become infinitesmal). So basically we solve $\frac{2+c_n}{1+c_n}=c_n$, which is $2+c_n=c_n+c_n^2$, so $c_n=\pm \sqrt{2}$. However, if $c_n$ is positive, then $c_{n+1}$ is positive. $c_1$ is positive, so it can't converge to $-\sqrt{2}$. Hence it converges to $\boxed{\sqrt{2}}$.

Answer 2

Hint:

$$\frac{a_{n+1}}{b_{n+1}}=1+\frac{1}{\frac{a_{n}}{b_{n}}+1}$$

So if $x$ serves indeed as limit for $\frac{a_n}{b_n}$ then it should satisfy...