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

rewrite as $$C_{n+1}  = 1 + \frac{C_n}{C_n + 1}= \frac{2C_n+1}{C_n + 1}=2-\frac{1}{1+C_n}$$
now
$$C_{n+2}  -C_{n+1}=2-\frac{1}{1+C_{n+1}}-(2-\frac{1}{1+C_{n}})=\\
\frac{1}{1+C_{n}}-\frac{1}{1+C_{n+1}}=
\\\frac{1+C_{n+1}-(1+C_{n})}{(1+C_{n+1})(1+C_{n})}=\\
\frac{C_{n+1}-C_{n}}{(1+C_{n+1})(1+C_{n})}=$$so now you can build your induction
$$ C_{n+1}>C_n \implies C_{n+2} > C_{n+1}$$because 
$$C_{n+2}  -C_{n+1}=\frac{C_{n+1}-C_{n}}{(1+C_{n+1})(1+C_{n})}=\frac{C_{n+1}-C_{n}}{(positive)(positive)}$$
