Cfrtjryk

We have the general term. With it, the existence and the value for the limit are proven.

For some values of $A$ and $B$ we have.

$$x_n=A(1-sqrt{2})^n+B(1+sqrt{2})^n$$

$$L=lim_{nto+infty}dfrac{x_n}{x_{n+1}}=lim_{nto+infty}dfrac{A(1-sqrt{2})^n+B(1+sqrt{2})^n}{A(1-sqrt{2})^{n+1}+B(1+sqrt{2})^{n+1}}$$

$$L=lim_{nto+infty}dfrac{dfrac{A(1-sqrt{2})^n}{(1+sqrt{2})^n}+dfrac{B(1+sqrt{2})^n}{(1+sqrt{2})^n}}{dfrac{A(1-sqrt{2})^{n+1}}{(1+sqrt{2})^n}+dfrac{B(1+sqrt{2})^{n+1}}{(1+sqrt{2})^n}}$$

$$L=dfrac{0+B}{0+B(1+sqrt{2})}=sqrt{2}-1$$

(the $0$'s come from the exponentials with base less that $1$, as they go to zero as the exponent goes to infinity)

No mind what values for $A$ and $B$ we have, so is, the limit is that, irrespective of the values for $x_1$ and $x_2$.

Added
Suppose the sequence has this form $x_n=A·r^n$ and check if it is possible for $r$ and $A$ to meet the conditions the recurrence law imposes:

$A·r^{n+1}=2A·r^n+A·r^{n-1}$ or $A·r^2r^{n-1}=A2·r·r^{n-1}+Ar^{n-1}$

But obviously $Aneq0$ and $rneq0$, so we can simplify and $r$ must satisfy:

$r^2-2r-1=0$ with roots $1-sqrt{2}$ and $1+sqrt{2}$

But the equation is linear, thus a linear combination of solutions is too a solution:

$x_n=A(1-sqrt{2})^n+B(1+sqrt{2})^n$

We have $x_n$ to be monotonically increasing and since $dfrac{x_{n+1}}{x_n}=2+dfrac{x_{n-1}}{x_n}$, we can say $dfrac{x_{n+1}}{x_n}<3$ as $dfrac{x_{n-1}}{x_n}<1$ (monotoniclly increasing)$,:forall:ngeq1$. So $x_n$ is convergent.

Let $x_n=Acdot a^n,:aneq0$. The recurrence relation becomes $$ Acdot a^{n+1}=2Acdot a^n+Acdot a^{n-1}implies Acdot a^{n-1}(a^2-2a-1)=0underset{substack{Aneq0\aneq0}}{implies} a^2-2a-1=0$$

By quadratic formula, $a=dfrac{2pmsqrt{4+4}}{2}=1pmsqrt{2}$. Since $x_n$'s are non negative, we have $a=displaystyle lim_{ntoinfty}x_n=1+sqrt{2}$
$rule{17cm}{0.5pt}$

$x_n=A(1+sqrt{2})^n+B(1-sqrt{2})^n$ where $A,B$ are independent of $n$ by linear recurrence.

So if $L=displaystylelim_{ntoinfty}dfrac{x_n}{x_{n+1}}=lim_{ntoinfty}dfrac{A(1+sqrt{2})^n+B(1-sqrt{2})^n}{A(1+sqrt{2})^{n+1}+B(1-sqrt{2})^{n+1}}=lim_{ntoinfty}dfrac{A+dfrac{B(1-sqrt{2})^n}{B(1+sqrt{2})^n}}{dfrac{A(1+sqrt{2})^{n+1}}{(1-sqrt{2})^n}+dfrac{B(1-sqrt{2})^{n+1}}{B(1+sqrt{2})^n}}=dfrac{A}{A(1+sqrt{2})}=sqrt{2}-1$.

$left(text{Since}: |1-sqrt{2}|<1implies displaystylelim_{ntoinfty}(1-sqrt{2})^nto0right)$.

To show that the limit $displaystylelim_{ntoinfty}dfrac{x_n}{x_{n+1}}$ exists we see that it is bounded since $$0<dfrac{x_n}{x_{n+1}}=dfrac{x_n}{2x_{n}+x_{n-1}}leqdfrac{x_n}{2x_{n}}=dfrac12 $$
and $x_{n+1}geq2x_nimplies x_{n+1}geq x_n$ which gives $dfrac{x_n}{x_{n+1}}=dfrac{2x_{n-1}+x_{n-2}}{2x_{n}+x_{n-1}}leqdfrac{2x_{n-1}+x_{n-2}}{2x_{n}}$ i.e.$dfrac{x_n}{x_{n+1}}-dfrac{x_{n-1}}{x_{n}}leqdfrac{x_{n-2}}{2x_{n}}$ which shows monotonicity.Hence the limit exists.

One trick is to use the given limit to derive the existence of it.

Write $y_n=frac{x_n}{x_{n+1}}$.

Claim: There exists $rin (0,1)$ such that
$$(sqrt 2-1)-y_nle rbig(y_{n-1}-(sqrt2-1)big).$$

Proof: Rearranging the terms,
begin{align*}
underbrace{frac{x_{n+1}}{x_n}}_{=1/ {y_n}}-1&=underbrace{frac{x_{n-1}}{x_n}}_{=y_{n-1}}+1\
frac 1{y_n}-(sqrt2+1)&=y_{n-1}-(sqrt2-1)\
frac 1{y_n}-frac1{sqrt2-1}&=y_{n-1}-(sqrt2-1)\
(sqrt 2-1)-y_n&=underbrace{y_n(sqrt2-1)}_{in(0,1)}big(y_{n-1}-(sqrt2-1)big).
end{align*}
so the claim is true. $square$

Therefore,
begin{align*}
|y_n-(sqrt2-1)|&le r |y_{n-1}-(sqrt2-1)|le r^2 |y_{n-2}-(sqrt2-1)|lecdots\
&le r^{n}|y_0-(sqrt2-1)|to0,
end{align*}
implies the limit exists, and $limlimits_{ntoinfty}y_n=sqrt2-1$.

Look at $x_n/x_{n+1} - x_{n-1}/x_n$ which is $(x_n^2 - x_{n-1}x_{n+1})/x_n x_{n-1}$.
We can prove by induction that the numerator is $(-1)^n$.
$$x_{n+1}^2 - x_n x_{n+2}
= x_{n+1}^2 - x_n(2x_{n+1} + x_n)
= x_{n+1}(x_{n+1} - 2x_n) - x_n^2
= -(x_n^2 - x_{n-1}x_{n+1})$$
with $x_1^2 - x_0x_2 = -1$. Hence $x_n/x_{n+1}$ tends to a limit by the alternating series test.

Since $x_n$ is increasing, all the terms become non-zero. By defining $a_n={x_nover x_{n+1}}$ we have $${x_nover x_{n+1}}={x_nover 2x_n+x_{n-1}}={1over 2+{x_{n-1}over {x_n}}}$$therefore $$a_n={1over 2+{ a_{n-1}}}$$Now by defining $b_n=a_n-(sqrt 2-1)$ we have $$b_n+sqrt 2-1={1over b_{n-1}+sqrt 2+1}$$therefore $$b_n=-sqrt 2+1+{1over b_{n-1}+sqrt 2+1}={(1-sqrt 2)b_{n-1}over a_n+2}$$since $x_n>0$ we have $a_n>0$ therefore$$|b_n|=|{(1-sqrt 2)b_{n-1}over a_n+2}|le {sqrt 2-1over 2}|b_{n-1}|$$which means that $b_n to 0$ or $a_n={x_nover x_{n+1}}to sqrt 2-1 quadblacksquare$

We have that

$$frac{x_{n+1}}{x_n} = 2 + frac{x_{n-1}}{x_n}$$

then by

$$y_n=frac{x_{n}}{x_{n-1}} implies y_{n+1}=2+frac1{y_n }quad y_0=2$$

which converges to $L=sqrt 2+1$ and then

$$lim_{nto infty} frac{x_n}{x_{n+1}}=lim_{nto infty} frac{1}{y_{n+1}}=frac1L=sqrt 2 -1$$

Let

$$lim_{nto infty} frac{x_n}{x_{n+1}} = lim_{nto infty} frac{x_{n-1}}{x_{n1}} = k$$

Now

$$lim_{nto infty} frac{x_n}{x_{n+1}} = k$$
$$lim_{nto infty} frac{x_n}{2x_{n} + x_{n-1}} = k$$

Take $x_n$ out from numerator and denominator
$$lim_{nto infty} frac{1}{2 + frac{x_{n-1}}{x_n}} = k$$

Using the first equation

$$ frac{1}{2 + k} = k$$
$$k^2+2k-1=0$$
This gives two solutions $k=sqrt{2}-1$ and $k=-sqrt{2}-1$. Since none of the terms can be negative, we reject . the second solution thereby giving us

$$lim_{nto infty} frac{x_n}{x_{n+1}} = k = sqrt2 - 1$$

EDIT - As suggested by the commenter we need to prove that it is a finite limit before we start with the proof. Initially it's a $frac{infty}{infty}$ form as both $x_n$ and $x_{n+1}$ approach $infty$ as $n$ approaches $infty$. I'll use a finite upper bound to show that the limit is finite which means it exists.

For any $n$

$$frac{x_n}{x_{n+1}} =frac{x_n}{2x_n + x_{n-1}}$$

As $x_{n-1}$ is always a positive quantity
$$frac{x_n}{x_{n+1}} leq frac{x_n}{2x_n}$$
$$frac{x_n}{x_{n+1}} leq frac{1}{2}$$

For any $n$, you can take the last statement to prove the monotonicity as
$$x_{n+1}geq x_n$$
And since both $x_n$ and $x_{n+1}$ are positive values, the lower bound is $0$. The upper bound along with lower bound and the monotonicity proves that the limit is finite.