Finite Continued Fraction Proof Let $\alpha = \left[a_0, a_1, a_2,\cdots,a_n\right]$ be a finite continued fraction with $a_0 > 0$ and let $C_i = p_i/q_i$ be the convergent of $\alpha$. 
If $i\ge 1$, prove that $\frac{p_i}{p_{i-1}} = \left[a_i, a_{i-1}, a_{i-2},\cdots, a_0\right]$ and $\frac{q_i}{q_{i-1}} = \left[a_i, a_{i-1}, a_{i-2},\cdots, a_1\right]$.
For the first equality, I want to use $p_i = a_i\cdot p_{i-1} + p_{i-2}$ to prove that $$\frac{p_i}{p_{i-1}} = a_i + \frac{1}{p_{i-1}/p_{i-2}}\;.$$ I do not know how to do it, though. 
 A: You already have:
$$\frac{p_i}{p_{i-1}}=a_i + \frac1{\dfrac{p_{i-1}}{p_{i-2}}}$$
Which is the induction step, in a proof by induction.

Complete procedure:
Let's prove by induction that, for $i\ge1$, then $\dfrac{p_i}{p_{i-1}}=[a_i,a_{i-1},\dots,a_0]$.  The analogous proof for $q_i$ is left to the reader.
Induction base:
First we test the induction base.  For $i=1$:
We know that $C_0=a_0=\frac{a_0}1$ which means $p_0=a_0$. 
Now: $C_1=a_0+\frac1{a_1}=\frac{a_0a_1+1}{a_1}$, so $p_1=a_0a_1+1$.  Therefor $\frac{p_1}{p_0}=\frac{a_0a_1+1}{a_0}=a_1+\frac1{a_0}=[a_1,a_0]$
Induction step:
If the preposition is true for $i=k$, then it must be true for $i=k+1$.
We know the identity: $p_{k+1}=a_{k+1}p_k+p_{k-1}$, so we divide by $p_k\ne0$ at both sides:
\begin{align}
p_{k+1} &= a_{k+1}p_k+p_{k-1} &&\text{divide by $p_k$:} \\
\frac{p_{k+1}}{p_k} &= a_{k+1}+\frac{p_{k-1}}{p_k} &&\text{Rewrting second summand:} \\
\frac{p_{k+1}}{p_k} &= a_{k+1}+\frac1{\dfrac{p_k}{p_{k-1}}} &&\text{By induction hypotes (true for $i=k$):} \\
\frac{p_{k+1}}{p_k} &= a_{k+1}+\frac1{[a_k,a_{k-1},\dots,a_0]} &&\text{By construction of continued fractions:} \\
\frac{p_{k+1}}{p_k} &= [a_{k+1},a_k,a_{k-1},\dots,a_0]
\end{align}
So it is proved by induction.
