\documentclass[12pt]{article}
\usepackage{amssymb,amsmath}
\pagestyle{empty}
\begin{document}

Problem 2. (1)(2) Identities. $(2\frac{1}{2})$ $q_n\geq n$. (3)(4) Odd terms and even terms converge. (5) The whole sequence converges. 

[Solution] (1)(2) We know that 

\[
\begin{pmatrix}
p_{k+1} & p_k\\ 
q_{k+1} & q_k
\end{pmatrix}
=
\begin{pmatrix}
p_k & p_{k-1}\\ 
q_k & q_{k-1}
\end{pmatrix}
\cdot
\begin{pmatrix}
a_k+1 & 1\\ 
1 & 0
\end{pmatrix}
\notag\\
=
\begin{pmatrix}
p_k a_{k+1} + p_{k-1} & p_k\\ 
q_k a_{k+1} + q_{k-1} & q_k
\end{pmatrix}
\]

This means 
\begin{align}
p_{k+1} &= p_k a_{k+1} + p_{k-1}\notag\\
q_{k+1} &= q_k a_{k+1} + q_{k-1}\notag
\end{align}
By computation, and these two formulae, we have 
\begin{align}
p_{k+1} q_k - q_{k+1} p_k &= - (p_k q_{k-1} - q_k p_{k-1})\notag\\
p_{k+1} q_{k-1} - q_{k+1} p_{k-1} &= a_{k+1} (p_k q_{k-1} - q_k p_{k-1})\notag
\end{align}
Inductively we get
\begin{align}
&p_n q_{n-1} - q_n p_{n-1} = (-1)^{n+1}\notag\\
&p_{n+2} q_n - q_{n+2} p_n = a_{n+2} (p_{n+1} q_n - q_{n+1} p_n) = a_{n+2} (-1)^n\notag
\end{align}

($2\frac{1}{2}$) $q_1 = a_1 q_0 + q_{-1} \geq 1\cdot 1 + 0 = 1$. $q_2 = a_2 q_1 + q_0 \geq 1\cdot 1 + 1 = 2$.  If $q_k\geq k$ and $q_{k+1} \geq k+1$, then
\begin{align}
q_{k+2} = a_{k+2} q_{k+1} + q_k \geq 1\cdot (k+1) + k \geq k+2\notag
\end{align}
Hence $q_n\geq n$. 

(3)(4) By the second part,
\begin{align}
\frac{p_{n+2}}{q_{n+2}} - \frac{p_n}{q_n} = (-1)^n \frac{a_{n+2}}{q_{n+2}q_n}\notag
\end{align}
Since $a_{n+2}$, $q_{n+2}$, and $q_n\in\mathbb{N}$, we prove that both sequence are decreasing and increasing respectively. By the first part we find 
\begin{align}
\frac{p_n}{q_n} - \frac{p_{n-1}}{q_{n-1}} = \frac{(-1)^{n+1}}{q_n q_{n-1}}\leq \frac{(-1)^{n+1}}{(n-1)^2}\qquad\qquad (\ast)\notag
\end{align}
If $k\textless l$, then 
\begin{align}
\frac{p_1}{q_1}\textgreater \frac{p_{2k+1}}{q_{2k+1}}\textgreater \frac{p_{2l+1}}{q_{2l+1}}\textgreater \frac{p_{2l}}{q_{2l}}\textgreater\frac{p_{2k}}{q_{2k}}\textgreater\frac{p_2}{q_2}\notag
\end{align}
Hence both sequences are bounded, and then convergent.

(5) By ($\ast$), we find that $\lim_{n\to\infty} \frac{p_{2k+1}}{q_{2k+1}} - \frac{p_{2k}}{q_{2k}} = 0 $. Since both limits exist, they are equal. Zipper theorem then yields that $(\frac{p_n}{q_n})$ converges.




\end{document}