Proof the convergence of the series $a_{n+1}=1-\frac{1}{2+a_{n}}$

Question

I try to improve my writing style and I went from "doesn't make any sense" and "wrong", to "missing lots of steps here", until "verbose". Can anyone make suggestions whether the proof is correct in the first step and how to improve it?

Prove that the series $a_{n+1}=1-\frac{1}{2+a_{n}}$ converges and determine the limit where $a_0=0$ and $a_n \in \mathbb{R}$

Step 0. We will use in the proof the folowing two facts

$a_n \leq b_n$ $\Rightarrow$ $lim \ a_n \leq lim \ b_n$ and $a_n \leq b_n \leq c_n$ and $lim \ a_n=lim \ c_n$$\Rightarrow$ $lim \ a_n = lim \ b_n = lim \ c_n$, where $a_n,b_n$ denote series in the real numbers $lim$ ist the limit and the theorem if a series is increasing monotonously and bounded then it converges and finally if $a_n=b_n$ then also $lim \ a_n = lim \ b_n$

We will first show that $0 \leq a_n \leq a_{n+1} \leq 1$ and then determine the limit.

Step 1 We first will define $f(x):=1-\frac{1}{2+x}$. Notice that $a_{n+1}=f(a_{n})$. If $0 \leq a \leq b $ then we also have $f(a) \leq f(b)$. We can conclude this by some simple rearrangements. Let's start with $$a \leq b$$ $$2+a\leq 2+b$$ $$-\frac{1}{2+a} \geq \frac{1}{2+b}$$ $$1-\frac{1}{2+a} \leq 1- \frac{1}{2+b}$$ and thus $$f(a) \leq f(b)$$

Step 2 We will show now that $0 \leq a_n \leq a_{n+1} \leq 1$

Induction start:

It is indeed the case that $0 \leq 0 \leq \frac{1}{2} \leq 1$

Induction step: We assume that it is indeed the case that $0 \leq a_n \leq a_{n+1} \leq 1$

We will use the Lemma which we have proved. It follows that $$\frac{1}{2} =f(0) \leq f(a_n) \leq f(a_{n+1}) \leq f(1)=\frac{2}{3}$$

But since $0 \leq \frac{1}{2}$ and $\frac{2}{3} \leq 1$ $f(a_n)=a_{n+1}$ and $f(a_{n+1})=a_{n+2}$

We can conclude that

$$0 \leq a_{n+1} \leq a_{n+2} \leq 1$$

which shows the induction step.

Step 3 Now we will show that the limit exists and determine the limit

Step 3.1 We will proof existence first. We know that the series is monotonously increasing and that it is bounded by $1$ thus it converges.

Step 3.2 We will determine the limit. Apply the limit on both sides of the equation $a_{n}=1-\frac{1}{2+a_n}$ and we find that $$ L = 1 - \frac{1}{2+L}$$. because if $a_n$ converges to a limit with $n>N$ then surely also $n+1>N$ converges with the same limit.

Then we find that the limit is the solution of the equation $L^2+L=1$. We know that $0 \leq L \leq 1$ because we also can interpret $0$ and $1$ as series and since from $a_n \leq b_n$ it also follows that $\lim a_n \leq \lim \ b_n$ and we find $0 \leq a_n \leq 1$.

We have two possible solutions now but one of them is negative and the only right answer is thus $L=\frac{\sqrt{5}}{2}-\frac{1}{2}$

Answer 1

Your proof is valid but informal and at some points, you could have skipped some steps.

You have used the word series instead of sequence.

Once you have introduced the function $$f(x):=1-\frac{1}{2+x}$$ It was easy to show that the function is increasing by taking derivative.

Overall it is clearly written proof.