I'm working on the following problem:
Let $\{ a_n \}_{n \geq 1}$ be a sequence of real numbers with $a_1 > -2$, $a_{n+1} = \sqrt{a_n + 2}$. Show that (i) $|a_{n+1} - 2| \leq \frac{1}{2} |a_n - 2|$, and (ii) find the limit of $a_n$ as $n \rightarrow \infty$.
My attempt:
First, we show that $a_{n+1} = \sqrt{a_n + 2} > 0$ via induction. Accordingly, when $n=1$,
$$ a_{2} = \sqrt{a_1+2} > \sqrt{-2+2} = 0. $$ Now, suppose $a_{k+1} > 0$ for some $k$. Then,
$$a_{k+2} = \sqrt{a_{k+1} + 2} >^{I.H.} \sqrt{2} > 0. $$ Thus, $a_{n+1} > 0$ for all $n$. Now, suppose $a_1 \neq 2$, and observe that \begin{align*} \frac{1}{2} |a_n - 2| &= \frac{1}{2} |(a_{n+1}^2 - 2) - 2| \\ &= \frac{1}{2}|(a_{n+1} - 2)(a_{n+1}+2)| \\ &> \frac{1}{2}(2) |a_{n+1} - 2| \qquad a_{n+1} > 0\\ &= |a_{n+1} - 2|. \\ \end{align*} Moreover, suppose $a_1 = 2$. Then, $\{ a_n \}$ is a constant sequence with $a_n = 2 \forall n$, so
$$ |a_{n+1} - 2| = 0 = \frac{1}{2}|a_n - 2|. $$
Combining these cases, it must hold that
$$ |a_{n+1} - 2| \leq \frac{1}{2} |a_n - 2|. $$
Concerning the limit, it can be shown that $\{ a_n \}$ is convergent, as a bounded monotone increasing sequence when $a_1 < 2$, a bounded monotone decreasing sequence when $a_1 > 2$, and as a constant sequence when $a_1 = 2$. We claim that $a_n \rightarrow 2$ as $n \rightarrow \infty$. Accordingly, let
$$ \lim_{n \rightarrow \infty} a_{n+1} = \lim_{n \rightarrow \infty} a_n = a. $$
Then, \begin{align*} a &= \lim_{n \rightarrow \infty} a_n \\ &= \lim_{n \rightarrow \infty} \bigg( a_{n+1}^2 - 2 \bigg) \\ &= a^2 - 2 \\ \Rightarrow 0 &= a^2 - a - 2 \\ \Rightarrow a &= -1, 2. \\ \end{align*}
Since the sequence is real-valued by assumption, it follows that $a_n \rightarrow 2$ (or, alternatively, $a_{n+1} > 0 \forall n \geq 1$).
As far as I can tell this seems like a valid solution, but I'm not proving (ii) in the way the writer of the question intended, as I don't use the result (i). I understand that result (i) tells us that each successive term has a distance from 2 that is less than or equal to half the distance of the previous term to 2, so this in itself seems to affirm that the limit must be 2. Is there a better way to approach this problem that utilizes this result?
Thanks!
Using the inequality you proved you have,
$$|a_{n+1}-2|<\frac{1}{2}|a_{n}-2|$$
You showed $\{a_n\}$ converges to some limit $a$ so taking limits gives
$$\lim\limits_{n\to\infty}|a_{n+1}-2|<\lim\limits_{n\to\infty}\frac{1}{2}|a_{n}-2|\\ |a-2|<\frac{1}{2}|a-2|\\ \implies |a-2|=0\\ \implies \lim\limits_{n\to\infty}a_n=2$$
You can also show $\{a_n\}$ converges, without using MCT, directly from the inequality. You can show by induction that $|a_{n}-2|<\frac{1}{2^{n-1}}|a_1-2|$ for $n\geq 2$ and the right hand side goes to $0$ as $n\to\infty$.
The nice thing about these two ways is you naturally avoid the artifact $\lim\limits_{n\to\infty}a_n=-1$, but solving for the limit using the recursive definition for $a_{n+1}$ is completely valid too.
edit: here is the induction proof that the limit exists. We first show that $|a_{n}-2|<\frac{1}{2^{n-1}}|a_1-2|$ for $n\geq 2$.
Base case: $n=2$. By the inequality you showed,
$$|a_2-2|<\frac{1}{2}|a_1-2|$$
Assume that $|a_{k}-2|<\frac{1}{2^{k-1}}|a_1-2|$ for some $k>2$. Then, by the inequality you showed,
$$|a_{k+1}-2|<\frac{1}{2}|a_k-2|<\frac{1}{2}\left(\frac{1}{2^{k-1}}|a_1-2|\right)=\frac{1}{2^{k}}|a_1-2|$$
and the induction proof is complete.
So, we have $|a_{n}-2|<\frac{1}{2^{n-1}}|a_1-2|$ for all $n\geq 2$. Taking limits on both sides gives,
\begin{align} \lim\limits_{n\to\infty}|a_{n}-2|&<\lim\limits_{n\to\infty}\frac{1}{2^{n-1}}|a_1-2|=0 \end{align}
By continuity of absolute value we have $\lim\limits_{n\to\infty}a_n=2$. We didn't assume the limit exists at any point.