Solution of A Recursively Defined Square-Root Sequence

Throughout we use that, for every $n\in\mathbb{N}$, the number $a_{n+1}$ satisfies $0\le a_{n+1}$ and $a_{n+1}^2=2+a_n$, which is what the definition of $\sqrt{2+a_n}$ says.

Step 1: $0\le a_n<2$ for every $n\in\mathbb{N}$. We argue by the principle of induction. For $n=1$ we have $a_1=1$, so $0\le a_1$ and $a_1<2$. Suppose $0\le a_n<2$. Then $0\le a_{n+1}$, and

$$a_{n+1}^2=2+a_n<2+2=4=2\cdot 2 .$$

Since $0\le a_{n+1}$ and $0\le 2$, clause 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field converts $a_{n+1}^2<2^2$ into $a_{n+1}<2$. Hence $0\le a_{n+1}<2$, completing the induction.

Step 2: $a_n\le a_{n+1}$ for every $n\in\mathbb{N}$. Fix $n$. By Step 1, $0\le a_n<2$, so $a_n-2<0$ and $0<1\le a_n+1$. By clause 4 of Elementary Order Arithmetic in an Ordered Field, $a_n-2<0$ gives $0<-(a_n-2)$; by clause 5 of that lemma, $0<\bigl(-(a_n-2)\bigr)(a_n+1)$; and since $\bigl(-(a_n-2)\bigr)(a_n+1)=-\bigl((a_n-2)(a_n+1)\bigr)$ by Elementary Arithmetic in an Ordered Field, clause 4 applied again yields

$$(a_n-2)(a_n+1)<0 .$$

Expanding the left side using Elementary Arithmetic in an Ordered Field gives $a_n^2-a_n-2<0$, that is,

$$a_n^2<2+a_n=a_{n+1}^2 .$$

Since $0\le a_n$ and $0\le a_{n+1}$, clause 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $a_n<a_{n+1}$; in particular $a_n\le a_{n+1}$.

Step 3: convergence. Let $A=\{a_n:n\in\mathbb{N}\}$. By Step 1 every element of $A$ is at most $2$, so $A$ is nonempty and bounded above, and by Step 2 the sequence satisfies $a_n\le a_{n+1}$ for every $n$. By clause 1 of A Bounded Monotone Sequence of Real Numbers Converges, $L=\sup A$ exists and $(a_n)$ converges to $L$. Since $2$ is an upper bound for $A$ and $L$ is the least upper bound, $L\le 2$; and since $a_1=1$ lies in $A$ and $L$ is an upper bound, $1\le L$.

Step 4: the shifted sequence converges to $L$ as well. Put $b_n=a_{n+1}$ for $n\in\mathbb{N}$. By the index-shift property recorded in The Real Line: Standing Notation and Background for Calculus on Intervals (vii), $(b_n)$ converges to $L$.

Step 5: identifying $L$. By clause 2 of Arithmetic of Limits of Real Sequences applied to the sequence $(b_n)$ and itself, the sequence $(b_n\,b_n)=(b_n^2)$ converges to $L\cdot L=L^2$. On the other hand, the constant sequence with value $2$ converges to $2$ by The Real Line: Standing Notation and Background for Calculus on Intervals (vii), so by clause 1 of Arithmetic of Limits of Real Sequences the sequence $(2+a_n)$ converges to $2+L$. But $b_n^2=a_{n+1}^2=2+a_n$ for every $n$, so the two sequences are equal, and by the uniqueness of limits in clause 1 of Uniqueness of Limits and Boundedness of Convergent Real Sequences,

$$L^2=2+L .$$

Rearranging with Elementary Arithmetic in an Ordered Field gives $L^2-L-2=0$, that is, $(L-2)(L+1)=0$. A field has no zero divisors, so $L-2=0$ or $L+1=0$, that is, $L=2$ or $L=-1$. By Step 3, $1\le L$, so $L\ne -1$. Therefore $L=2$, that is,

$$\lim_{n\to\infty}a_n=2 .$$