Prove $\lim_{n \rightarrow \infty} \frac{u_{n+1}}{u_{n}} = C,$ with $u_{0}=1$ and $u_{1}=2$. I have the following recurrence relation
\begin{align}
    u_{0} &= 1 \nonumber \\
    u_{1} &= 2 \nonumber \\
    u_{n} &= u_{n-2} + u_{n-1} +1
\end{align}
Now I want to prove that the ratio after $n$ (where $n$ is large) remains constant. I can see this by simple putting the equation in excel and then see that the ratio grows to the Golden Ratio (i.e. $\frac{1+\sqrt{5}}{2}$).
Hence I want to prove $$\lim_{n \rightarrow \infty} \frac{u_{n+1}}{u_{n}} = C,$$
for some $C \in \mathbb{R}$ (more specifically $C = \frac{1+\sqrt{5}}{2}$.
I find
$$\frac{u_{n+1}}{u_{n}} = \frac{u_{n-2}+u_{n-1}+1}{u_{n-1}+u_{n}+1},$$
but this does not really help me. I would say we can ommit the $+1$ in the equations since $u_{n}$ grows large but this is not really mathematically sound.
Any suggestions are more than welcome.
 A: Define the sequence $(v_n)_{n \in \mathbb{N}}$ by $v_n:=u_n+1$ so that $v_0=2, v_1=3, v_n = v_{n-1}+v_{n-2}$ for all $n \geq 2$. Then an easy induction argument yields $v_n = F_{n+3}$ where $F_n$ denotes the $n$-th term of the Fibonacci sequence (defined by $F_0:=0, F_1:=1$ and $F_n = F_{n-1}+F_{n-2} \forall n \geq 2$), whence for all $n \geq 0$ we have $u_n = F_{n+3}-1 = \frac{1}{\sqrt{5}} (\alpha^{n+3} - \beta^{n+3})$ where $\alpha = \frac{1+\sqrt{5}}{2}, \beta = \frac{1-\sqrt{5}}{2}$, using the well-known Binet's Formula (which can be derived simply by looking at the characteristic equation of the Fibonacci sequence). This finally yields 
$$\lim_{n \rightarrow \infty} \frac{u_{n+1}}{u_n} = \lim_{n \rightarrow \infty} \frac{F_{n+4}-1}{F_{n+3}-1} = \lim_{n \rightarrow \infty} \frac{F_{n+1}-1}{F_n-1} = \lim_{n \rightarrow \infty} \frac{\alpha^{n+1} - \beta^{n+1} - \sqrt{5}}{\alpha^n - \beta^n - \sqrt{5}} = \alpha \left( \lim_{n \rightarrow \infty} \frac{1 - \lambda^{n+1} - \frac{\sqrt{5}}{\alpha^{n+1}}}{1 - \lambda^n - \frac{\sqrt{5}}{\alpha^n}} \right) = \alpha = \frac{1+\sqrt{5}}{2}$$
where we have used $\beta<1<\alpha \implies \lambda := \frac{\beta}{\alpha}<1 \implies \lim_{n \rightarrow \infty} \lambda^n=0$.
A: Note that we have $\frac{u_{n+1}}{u_n}<2 $ and the sequence is monotonically increasing,  and bounded above,  it guarantees the existence of the limit. 
