Prove that the sequence $C_{n+1} = 1 + \frac{C_n}{C_n + 1}$ is monotonically increasing by induction I have a sequence
$$C_{n+1}  = 1 + \frac{C_n}{C_n + 1}$$
With a base case $C_1 = 3/2$ and want to prove that it's monotonically increasing by induction.
Whenever I try to prove $C_n < C_{n+1} \implies C_{n+1} < C_{n+2}$ 
where $$C_{n+2} = 1 + \frac{2C_n + 1}{3C_n + 2} $$
I get to the inequality $$ \frac{C_n}{C_n + 1} < \frac{2C_n + 1}{(C_n + 1)^2} $$
but don't know how to get to the point where that implies that $C_{n+1} < C_{n+2}$
Thank

By induction, it is clear that $C_n>0$ for all $n$. Put $c=C_n$.
You want : $C_{n+1}\gt C_n$, which is equivalent to
$$
\begin{array}{lcl}
C_{n+1}\gt C_n & \Leftrightarrow & 1+\frac{C_n}{C_n+1} \gt C_n \\
  & \Leftrightarrow & 1+\frac{c}{1+c} \gt c \\
  & \Leftrightarrow & \frac{1+2c}{1+c} \gt c \\
  & \Leftrightarrow & 1+2c \gt c+c^2 \\
  & \Leftrightarrow & 0 \gt -1-c+c^2 \\
  & \Leftrightarrow & 0 \gt \big(c-\frac{1}{2}\big)^2-\frac{5}{4}\\
  & \Leftrightarrow & \frac{5}{4} \gt \big(c-\frac{1}{2}\big)^2\\
  & \Leftrightarrow & \frac{-\sqrt{5}}{2} \lt c-\frac{1}{2} \lt \frac{\sqrt{5}}{2}\\
  & \Leftrightarrow & \frac{1-\sqrt{5}}{2} \lt c \lt \frac{1+\sqrt{5}}{2}\\
\end{array}
$$
Put $\alpha=\frac{1-\sqrt{5}}{2}$ and $\beta=\frac{1+\sqrt{5}}{2}$.
So, the goal is now to show that $\alpha \lt C_n \lt \beta$ for every $n$.
Can you finish from here ? (use induction and $C_{n+1}=2-\frac{1}{1+C_n}$).
