How do I show that if $p_n q_{n-1} - p_{n-1}q_n = 1$, then $p_n/q_n$ converges? Let $p_{n}$ and $q_{n}$ be strictly increasing, integer valued, sequences. Show that if
$$p_{n}q_{n-1}-p_{n-1}q_{n}=1,$$
for each integer $n \ge 1$, then the sequence of quotients $\frac{p_{n}}{q_{n}}$ converges.
My attempt: For given $\epsilon >0$, there exists a positive integer $N$ such that for $m \geq n \geq N$, and $\frac{1}{q_{i}} \neq 0$ for $i=1,2,3,\dotsc$
\begin{align}
\left| \frac{p_{n}}{q_{n}} - \frac{p_{m}}{q_{m}} \right|
&=\left| \frac{p_{m}}{q_{m}} - \frac{p_{n}}{q_{n}} \right| \\
&\leq \left| \frac{p_{m}}{q_{m}} - \frac{p_{m-1}}{q_{m-1}} \right| + \left|\frac{p_{m-1}}{q_{m-1}} - \frac{p_{m-2}}{q_{m-2}} \right| +\dotsb +\left|\frac{p_{n+1}}{q_{n+1}} - \frac{p_{n}}{q_{n}} \right| \\
&\leq \left| \frac{1}{q_{m}q_{m-1}} \right| + \left| \frac{1}{q_{m-1}q_{m-2}} \right|+\dotsb+ \left| \frac{1}{q_{n+1}q_{n}} \right| \\
&\leq \frac{1}{q_{m-1}^{2}} + \frac{1}{q_{m-2}^{2}}+\dotsb+\frac{1}{q_{n}^{2}} \\
&= \sum_{i=n}^{m-1} \frac{1}{q_{i}^{2}} \\
& \leq \sum_{k=1}^{\infty} \frac{1}{k^{2}}.
\end{align}
We know that $\sum_{k=1}^{\infty} \frac{1}{k^{2}}$ is convergent. This implies $\sum_{k=n}^{m} \frac{1}{k^{2}} \leq \epsilon$ if $m \geq n \geq N$.  This implies $|\frac{p_{n}}{q_{n}} - \frac{p_{m}}{q_{m}}|\leq \epsilon$ for all $m \geq n \geq N$.  Hence, $\frac{p_{n}}{q_{n}}$ converges.
Can anyone please suggest me any improvement to justify this claim?
Is my last few steps are correct, where I apply the comparison test?

The question is tagged proof-writing, hence this answer is intended to address the style of the proof given, rather than to offer a more slick or elegant proof.  My general critiques are as follows:


*

*The introduction of $\varepsilon$ is somewhat confusing.  The argument presented relies on the fact that the sequence of partial sums
$$ S_n = \sum_{k=1}^{n} \frac{1}{k^2} $$
is Cauchy which means that for any $\varepsilon > 0$, there exists an $N$ so large that $n > m > N$ implies that
$$ |S_n - S_m| = \sum_{k=m+1}^{n} \frac{1}{k^2} < \varepsilon. $$
Thus I would start the proof by fixing $\varepsilon$ and choosing $m$ and $n$ which get this job done.  

*The statement of the exercise gives several hypotheses.  The argument given does not explicitly invoke these hypotheses at any point, and uses them in a way which require some computation.  While this computation is routine, it detracts from the clarity of the argument.  It is good style to note when hypotheses are being used, particularly when they are being used in a way that is not immediately obvious.  Remember that the goal of a proof is not to show off your knowledge or ability, but to clearly communicate an idea to the reader.

*In the first step of the computation, the writer states
$$ \left| \frac{p_n}{q_n} - \frac{p_m}{q_m} \right|
= \left| \frac{p_m}{q_m} - \frac{p_n}{q_n} \right|.$$
I understand the reasoning behind this—the author wants the indexing to run in the "correct" order.  However, this step is completely unnecessary.  Skip routine computation when it won't interfere with the reader's comprehension.
This seems to contradict the previous point.  I will claim that it doesn't, really.  There is a balance between saying too much and saying too little.  The precise boundary between the two is a matter of taste, but I would say that properties of the absolute value of a difference are "obvious", while the use of hypotheses in an argument should be highlighted.

*This is very much a matter of personal style, and one can have a difference of opinion without being wrong, but I (personally) don't really like ellipses in proofs.  They are used to hide induction arguments in a way that I find vaguely off-putting.  This isn't an absolute rule and is entirely a matter of personal choice, but I prefer to avoid ellipses unless it has a negative impact on readability.  In this case, I think that sigma notation can eliminate the ellipses without hurting the clarity of the argument.

*The given proof shows that the sequence $\{p_n/q_n\}$ is Cauchy.  Cauchy sequences are not a priori convergent—it is only in a complete metric space that Cauchy implies convergent.  I think that it is worth spending a couple of words to point this out.

The following is my (rough, unedited) presentation of the original proof:
Proposition:  Let $p_{n}$ and $q_{n}$ be strictly increasing, integer valued, sequences. Show that
$$p_{n}q_{n-1}-p_{n-1}q_{n}=1,$$
for each integer $n \ge 1$, then the sequence of quotients $\frac{p_{n}}{q_{n}}$ converges.
Proof: Fix $\varepsilon > 0$.  The sequence
$$ S_n = \sum_{k=1}^{n} \frac{1}{k^2} $$
is Cauchy, hence there is an $N$ so large that if $m$ and $n$ are integers with $n \ge m \ge N$, then
$$ |S_n - S_m| = \sum_{k=m+1}^{n} \frac{1}{k^2} < \varepsilon. $$
Then for any such $m$ and $n$,
\begin{align}
\left| \frac{p_n}{q_n} - \frac{p_m}{q_m} \right|
&= \left| \frac{p_n}{q_n} + \sum_{k=m+1}^{n-1} \left( \frac{p_k}{q_k} - \frac{p_k}{q_k} \right) - \frac{p_m}{q_m} \right| \\
&\le \sum_{k=m+1}^{n} \left| \frac{p_k}{q_k} - \frac{p_{k-1}}{q_{k-1}} \right|.
   && (\text{triangle inequality})
\end{align}
By hypothesis,
$$ p_n q_{n-1} - p_{n-1} q_{n} = 1
\implies \frac{p_n}{q_{n}} - \frac{p_{n-1}}{q_{n-1}} = \frac{1}{q_n q_{n-1}}
$$
for all $n \in \mathbb{N}$.  Thus
\begin{align}
\sum_{k=m+1}^{n} \left| \frac{p_k}{q_k} - \frac{p_{k-1}}{q_{k-1}} \right|
&= \sum_{k=m+1}^{n} \left| \frac{1}{q_n q_{n-1}} \right| \\
&\le \sum_{k=m+1}^{n} \frac{1}{q_{n-1}^2}.
     && (\text{$q_n > 0$ is a strictly increasing sequence})
\end{align}
As $q_n$ is a strictly increasing sequence of integers, it follows that $q_{n-1} \ge n$ for each integer $n \ge 1$, therefore
\begin{align}
\left| \frac{p_n}{q_n} - \frac{p_m}{q_m} \right|
\le \sum_{k=m+1}^{n} \frac{1}{q_{n-1}^2}
\le \sum_{k=m+1}^{n} \frac{1}{k^2}
< \varepsilon.
\end{align}
This demonstrates that the sequence $\{p_n/q_n\}$ is a Cauchy sequence in $\mathbb{R}$.  Since $\mathbb{R}$ is complete, this sequence converges in $\mathbb{R}$.
