The sequence $u_n$ is defined by $u_0 = 0, u_1 =1, u_n = 3u_{n-1}+u_{n-2}$. Calculate $\lim_{n\rightarrow \infty}u_n/u_{n-1}$ The sequence $u_n$ is defined by $u_0 = 0, u_1 =1, u_n = 3u_{n-1}+u_{n-2}$. Calculate $\lim_{n\rightarrow \infty}u_n/u_{n-1}$


The Solution:

$u_n$ is given by a linear combination of powers $\lambda_1^n, \lambda_2^n$ of $\lambda^2=3\lambda+1$ (with nonzero coefficients), so the limit of the ratio $u_n/u_{n-1}$ is the root $(3+\sqrt{13})/2$ of largest absolute value.

Question - without looking at the solution, I would set it up as a matrix equation 

\begin{gather}
 \begin{bmatrix} u_{n} \\u_{n-1} \end{bmatrix}
 =
  \begin{bmatrix}
   3 &amp;
   1 \\
   1 &amp;
   0 
   \end{bmatrix}
 \begin{bmatrix} u_{n-1} \\u_{n-2} \end{bmatrix}
 =
  \begin{bmatrix}
   3 &amp;
   1 \\
   1 &amp;
   0 
   \end{bmatrix}^{n-1}
 \begin{bmatrix} u_{1} \\u_{0} \end{bmatrix}
 =
  \begin{bmatrix}
   3 &amp;
   1 \\
   1 &amp;
   0 
   \end{bmatrix}^{n-1}
 \begin{bmatrix} 1 \\0 \end{bmatrix}
\end{gather}

and then I can diagonalize   

\begin{gather}
A = \begin{bmatrix}
   3 &amp;
   1 \\
   1 &amp;
   0 
   \end{bmatrix} = PDP^{-1}
\end{gather}
where D is the diagonal matrix with the entries $\lambda = (3 \pm \sqrt{13})/2$, solve for $u_n$ and $u_{n-1}$ and presumably reach the same conclusion.

Can someone explain the given solution to me? I see that $u_n$ is given by a linear combination of powers $\lambda_1^n, \lambda_2^n$ of $\lambda^2=3\lambda+1$ (with nonzero coefficients), but not sure how they get to the conclusion.
For such a linear recurrence, the general solution is $\newcommand{\la}{\lambda}a\la_1^n+b\la_2^n$. Here $\la_1&gt;|\la_2|$ and so, if $a\ne0$
$$\frac{u_{n+1}}{u_n}=\frac{a\la_1^{n+1}+b\la_2^{n+1}}
{a\la_1^n+b\la_2^n}
=\la_1\frac{1+(b/a)(\la_2/\la_1)^{n+1}}{1+(b/a)(\la_2/\la_1)^{n}}.
$$
As $|\la_2/\la_1|&lt;1$ this tends to $\la_1$. 

Of course, you may be unlucky and have a particular solution with $a=0$,
in which case this argument fails. But here, you don't.
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$


  Following your result:


\begin{align}
&amp;\text{As}\quad n \to \infty\,,
\\ &amp;\
{u_{n} \choose u_{n - 1}} = \pars{\begin{array}{cc}\ds{3} &amp; \ds{1} \\ \ds{1} &amp; \ds{0}\end{array}}^{n - 1}
{1 \choose 0}
\\[2mm] &amp;\ \sim
\pars{3 + \root{13} \over 2}^{n - 1}
{{3 + \root{13} \choose 2} \over \root{\pars{3 + \root{13}}^{2} + 4}}\,
{\pars{3 + \root{13} \quad 2} \over \root{\pars{3 + \root{13}}^{2} + 4}}
{1 \choose 0}
\\[2mm] &amp;\ \propto
\pars{3 + \root{13} \over 2}^{n - 1}
{3 + \root{13} \choose 2}
\implies
\bbx{\lim_{n \to \infty}{u_{n} \over u_{n - 1}} = {3 + \root{13} \over 2}}
\approx 3.3028
\end{align}