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Show that, for the recurrence $$a_n=2a_{n-1}+a_{n-2},$$ the ratio $\dfrac{a_{n}}{a_{n-1}}$ tends to $\sqrt{2}-1$ as $n\to\infty$.

Not sure about this one, I know the recurrence relation solution is $$a_n=\frac{1}{2}\left((1-\sqrt{2})^n+(1+\sqrt{2})^n\right)$$ but not sure how to proceed from here.

Blue
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thesmallprint
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    How do you know $a_n=\frac{1}{2}\Big((1-\sqrt{2})^n+(1+\sqrt{2})^n\Big)$? Perhaps you know $a_0$ and $a_1$? Otherwise you know $a_n=c(1-\sqrt{2})^n+d(1+\sqrt{2})^n$ for some $c$ and $d$, which so long as $d\not=1$ should lead to the ratio tending towards .$(1+\sqrt{2})$ – Henry Aug 28 '20 at 08:28
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    So if $a_0=a_1=1$ then you get the sequence $1,1,3,7,17,\ldots$ with the ratio tending towards $1+\sqrt{2}$, while if $a_0=1,a_1=1-\sqrt{2}$ you get the sequence $1, 1-\sqrt{2},3-\sqrt{8},7-\sqrt{50}, 17-\sqrt{288}, \ldots$ with the ratio tending towards $1-\sqrt{2}$, neither of which are your $\sqrt{2}-1$ – Henry Aug 28 '20 at 08:35
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    and my initial comment should have said "so long as $d \not = 0$" – Henry Aug 28 '20 at 08:37

3 Answers3

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We will assume that $a_0, a_1 > 0$.

Let $b_n=a_n/a_{n-1}$. You have $b_n=2+1/b_{n-1}$. Noting $x$ the limit of $b_n$ you get $x=2+1/x$. Can you conclude from here ?

By the way, the limit is $1+\sqrt{2}$ and not $\sqrt{2}-1$.

EDIT: The fact that $b_n$ admits a limit is immediate since $(b_n)$ is bounded in $[2,2.5]$ from rank $3$, with a unique accumulation point.

fonfonx
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    This is true, as long as you first show that the limit exists (and so you can use limit arithmetic). – GSofer Aug 28 '20 at 08:26
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If you already know that $$ a_n=\frac{1}{2}\big((1-\sqrt{2})^n+(1+\sqrt{2})^n\big), $$ then you can directly calculate that $$ \frac{a_n}{a_{n-1}}=\frac{(1-\sqrt{2})^n+(1+\sqrt{2})^n}{(1-\sqrt{2})^{n-1}+(1+\sqrt{2})^{n-1}}; $$ multiplying numerator and denominator by $(1+\sqrt{2})^{-n}$ (the reciprocal of the biggest term in sight) yields $$ \frac{a_n}{a_{n-1}}=\frac{(1-\sqrt{2})^n(1+\sqrt{2})^{-n}+1}{(1-\sqrt{2})^{n-1}(1+\sqrt{2})^{-n}+(1+\sqrt{2})^{-1}}. $$ Can you find the limit of this expression as $n\to\infty$?

Greg Martin
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Let $t=1+\sqrt 2$ and $s=\sqrt 2-1$. Then $\frac {a_n} {a_{n-1}}=\frac {t^{n}+s^{n}} {t^{n-1}+s^{n-1}}=\frac {t^{n}(1+u^{n})} {t^{n-1}(1+u^{n-1})} $ where $u=\frac s t$. Since $0<u <1$ we get $u^{n} \to 0$. Hence the limit is $t=1+\sqrt 2$.

Kavi Rama Murthy
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