<s>[INST] How to prove that $a_{n+1} = 1 + \frac{a_n}{a_n +1}$ is monotone increasing with $a_1=1$ This question has been driving me crazy, and I can't find the answer anywhere. 
I tried proving it using induction. 
As for the base case:
$a_1=1<1.5=a_2$
Next, suppose $a_k \leq a_{k+1}$
Then, 
$$a_k \leq a_{k+1}$$
$$a_k +1 \leq a_{k+1}+1$$
$$ \frac{1}{a_k+1}\geq \frac{1}{ a_{k+1}+1} $$
$$ \frac{a_k}{a_k+1}\geq \frac{a_k}{ a_{k+1}+1} $$
$$ \frac{a_k}{a_k+1} +1\geq \frac{a_k}{ a_{k+1}+1}+1 $$
My goal was to get 
$$ \frac{a_k}{a_k+1} +1\leq \frac{a_{k+1}}{ a_{k+1}+1}+1 $$
$$a_{k+1} \leq a_{k+2}$$
But obviously, I was not able to find a way. 
I also tried a different inductive approach where I noted 
$$a_{k+2}=\frac{5 a_n +3}{3 a_n +2}$$
You can check that this is true by seeing you get the correct $a_3=1.6$ using the given $a_1=1$. 
Assuming $a_k \leq a_{k+1}$ I had that 
$$ a_k \leq 1 + \frac{a_n}{a_n +1}$$
$$ \frac{a_n}{a_n +1} +1 \leq \frac{{a_n}^2 + 4a_n +2}{{a_n}^2+2a_n+1}$$
From the assumption $a_k \leq 1 + \frac{a_n}{a_n +1}$, we have that ${a_n}^2 \leq a_n +1$, so 
$$ \frac{a_n}{a_n +1} +1 \leq \frac{{a_n}^2 + 4a_n +2}{{a_n}^2+2a_n+1} \leq \frac{(a_n+1) +4a_n +2}{{a_n}^2+2a_n+1} =\frac{5a_n+3}{{a_n}^2+2a_n+1}$$
If only I could change the denominator the same way while keeping the direction of the inequality, I could substitute $a_n+1$ for ${a_n}^2$ and I would get the desired $\frac{5 a_n +3}{3 a_n +2}$, proving that $$a_{k+1} \leq a_{k+2}$$ since $$a_{k+2}=\frac{5 a_n +3}{3 a_n +2}$$. 
How do I prove the sequence is monotonic increasing? Both of these routes did not get me to the answer. 
 [/INST] You could also show that $1 + \frac{a_i}{a_i+1} > a_i$, by rearranging the inequality to get $2a_i+1>a_i^2+1$ then $2>a_i$. By showing $a_{n+1} = 2 - \frac{1}{1+a_n}$ (as in earlier answer) and that $a_i>0$, you show that $2>a_i$.
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