How would I solve the following problem on the limit of a sequence?

Question

I am a little confused as to how I can compute the limit for this. I know the sequence comes out to be $x^2 = 2x + 1$ but I am not sure how to proceed further.

Any help?

Answer 1

If $$a_{n+1}=2a_{n}+a_{n-1}$$ then $${a_{n+1}\over a_n} = 2+{1\over {a_{n}\over a_{n-1}}}$$

Write: $b_n = {a_{n}\over a_{n-1}}$ then we have $$ b_{n+1}= 2+{1\over b_n}$$ Let $b= \lim_{n\to \infty} b_n$ (here you must prove first that $b$ exist) then we have to solve: $$b = 2+{1\over b}$$ and we get $b^2-2b-1=0$ so $b= 1\pm \sqrt{2}$...

Answer 2

From solving the characteristic equation, you know that the general solution of the recurrence is of the form

$$a_n=c_0r_0^n+c_1r_1^n$$ where $r_0,r_1$ are the roots.

Then, assuming $|r_0|>|r_1|$,

$$\frac{a_{n+1}}{a_n}=\frac{c_0r_0^{n+1}+c_1r_1^{n+1}}{c_0r_0^n+c_1r_1^n}=\frac{r_0+r_1\dfrac{c_1}{c_0}\left(\dfrac{r_1}{r_0}\right)^n}{1+\dfrac{c_1}{c_0}\left(\dfrac{r_1}{r_0}\right)^n}$$ tends to $r_0$, provided $c_0\ne0$