How to prove that $a_{n+1} = 1 + \frac{a_n}{a_n +1}$ is monotone increasing with $a_1=1$

Question

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.

Answer 1

Hint: $\,a_{n+1} = 2 - \cfrac{1}{a_n +1}\,$, so $\,a_{n+1}-a_n=\cfrac{1}{a_{n-1} +1}-\cfrac{1}{a_n +1}=\cfrac{a_n-a_{n-1}}{(a_n +1)(a_{n-1} +1)}\,$.

Answer 2

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 $2>a_i$.