The Recurrence x(n+1)=sqrt(2+x(n))

Introduction

A nested radical such as \(\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}\) suggests that a number may be repeated indefinitely inside itself. To make that idea precise, define each new term from the preceding one:

x_{n+1}=\sqrt{2+x_n}.

The question is not merely what number might remain unchanged under this rule. We must also prove that the terms actually approach that number. The proof depends on the starting value: below 2 the terms increase, above 2 they decrease, and at 2 they stay fixed. In all three cases, an invariant bound keeps the terms on the appropriate side of 2.

Definition and the candidate limit

Definition: Choose a positive initial value \(x_1>0\). Define the sequence \((x_n)\) recursively by \(x_{n+1}=\sqrt{2+x_n}\) for every \(n\in\mathbb{N}\).

Every term is positive: if \(x_n>0\), then \(2+x_n>0\), so its square root is positive. Thus the recurrence is defined at every index. A number \(L\) that remains unchanged when the recurrence is applied must satisfy \(L=\sqrt{2+L}\). Since the square root is nonnegative, a possible such number must be nonnegative. Squaring gives

L^2=L+2 \qquad\Longleftrightarrow\qquad (L-2)(L+1)=0.

The algebraic candidates are 2 and \(-1\), but \(-1\) cannot satisfy the original equation: its right-hand side would be \(\sqrt{1}=1\). Therefore 2 is the only fixed point of the recurrence on the nonnegative real numbers. This identifies a candidate, not yet a proof of convergence.

Worked terms from different starting values

Worked Example: Starting below 2

Let \(x_1=1\). Apply the recurrence repeatedly, retaining exact expressions:

\begin{aligned} x_1&=1,\\ x_2&=\sqrt{2+x_1}=\sqrt{2+1}=\sqrt{3},\\ x_3&=\sqrt{2+x_2}=\sqrt{2+\sqrt{3}},\\ x_4&=\sqrt{2+x_3}=\sqrt{2+\sqrt{2+\sqrt{3}}}. \end{aligned}

Each displayed term checks directly in the defining recurrence: for instance, \(x_3=\sqrt{2+x_2}\), and \(x_4=\sqrt{2+x_3}\). These terms lie below 2 after the first term: \(x_2<2\) because \(3<4\), and \(x_3<2\) because \(2+\sqrt{3}<4\), as \(\sqrt{3}<2\). The general argument below proves that the terms increase while remaining below 2.

Worked Example: Starting above 2

Let \(x_1=3\). The first terms are

\begin{aligned} x_1&=3,\\ x_2&=\sqrt{2+x_1}=\sqrt{2+3}=\sqrt{5},\\ x_3&=\sqrt{2+x_2}=\sqrt{2+\sqrt{5}},\\ x_4&=\sqrt{2+x_3}=\sqrt{2+\sqrt{2+\sqrt{5}}}. \end{aligned}

The recurrence checks at each step by substituting the preceding expression under the square root. Also \(2<\sqrt{5}<3\), since \(4<5<9\). To check that \(x_3< x_2\), both are positive, so compare their squares: \(x_3^2=2+\sqrt{5}<5=x_2^2\), because \(\sqrt{5}<3\). Thus the first step moves downward but remains above 2. The proof below establishes these properties for every term.

Worked Example: Starting at the fixed point

Let \(x_1=2\). Then

x_2=\sqrt{2+2}=2,\qquad x_3=\sqrt{2+2}=2,\qquad x_4=\sqrt{2+2}=2.

Substitution confirms that \(2\) reproduces itself under the recurrence. Inductively, if \(x_n=2\), then \(x_{n+1}=\sqrt{2+2}=2\). Hence every term equals 2, and the sequence converges to 2.

An invariant interval

The map governing the recurrence is \(g(x)=\sqrt{2+x}\), defined for \(x\geq -2\). On nonnegative inputs it is order-preserving: if \(u\leq v\), then \(2+u\leq 2+v\), and taking nonnegative square roots preserves that inequality. This lets us establish an interval that contains every term.

Invariant interval: An interval \(I\) is invariant under a map \(g\) if \(x\in I\) implies \(g(x)\in I\). If the initial value lies in such an interval, every subsequent term generated by the recurrence remains in it.

The interval is chosen according to \(x_1\). If \(0<x_1<2\), use \([x_1,2]\). For any \(x\) in this interval, order preservation gives

\sqrt{2+x_1}\leq g(x)\leq \sqrt{4}=2.

Moreover, \(g(x_1)>x_1\). To verify this, both quantities are positive, and

g(x_1)>x_1 \quad\Longleftrightarrow\quad 2+x_1>x_1^2 \quad\Longleftrightarrow\quad (2-x_1)(x_1+1)>0,

which holds for \(0<x_1<2\). Therefore \(g(x)\geq g(x_1)>x_1\), while \(g(x)\leq2\); the interval maps into itself.

If \(x_1>2\), use \([2,x_1]\). For \(x\) in this interval, order preservation gives \(2\leq g(x)\leq g(x_1)\). Also \(g(x_1)<x_1\), since

g(x_1)<x_1 \quad\Longleftrightarrow\quad 2+x_1<x_1^2 \quad\Longleftrightarrow\quad (x_1-2)(x_1+1)>0.

Thus \(g(x)\) remains in \([2,x_1]\). If \(x_1=2\), the interval consisting only of 2 is invariant. In every case the recurrence remains within a bounded interval.

Monotonicity on either side of 2

To compare consecutive terms, subtract \(x_n\) from \(x_{n+1}\) and rationalize. The denominator is positive because both terms are positive:

\begin{aligned} x_{n+1}-x_n &=\sqrt{2+x_n}-x_n\\ &=\frac{(2+x_n)-x_n^2}{\sqrt{2+x_n}+x_n}\\ &=\frac{(2-x_n)(x_n+1)}{\sqrt{2+x_n}+x_n}. \end{aligned}

Since \(x_n+1>0\), the sign of the difference is the sign of \(2-x_n\). The invariant intervals already show that terms starting below 2 remain below 2, and terms starting above 2 remain above 2. Consequently, the difference is positive in the first case and negative in the second.

Conclusion: If \(0<x_1<2\), the sequence is strictly increasing and bounded above by 2. If \(x_1>2\), it is strictly decreasing and bounded below by 2. If \(x_1=2\), it is constant.

Why the sequence converges and what its limit is

The Monotone Convergence Theorem states that every increasing sequence of real numbers that is bounded above converges to a real limit; likewise, every decreasing sequence bounded below converges. The interval bounds and monotonicity just proved meet these conditions. Thus the sequence has a finite limit \(L\) for every positive initial value.

It remains to identify that limit. The square-root function is continuous on its domain, and all terms are positive. Taking limits on both sides of the recurrence is therefore valid:

\lim_{n\to\infty}x_{n+1} = \lim_{n\to\infty}\sqrt{2+x_n} = \sqrt{2+\lim_{n\to\infty}x_n}.

A shifted convergent sequence has the same limit, so the left side is \(L\). Hence \(L=\sqrt{2+L}\). Since \(L\geq0\), squaring and factoring gives

L^2-L-2=0, \qquad (L-2)(L+1)=0, \qquad L=2.

The root \(-1\) is excluded by \(L\geq0\), as well as by the original square-root equation. The argument proves the full result: no matter which positive value begins the recurrence, the terms converge to 2.

Choose the bound.
Use \([x_1,2]\) below 2, \([2,x_1]\) above 2, or the single point 2 when \(x_1=2\).

Keep the terms in that interval.
The recurrence map is order-preserving, and the selected interval maps into itself.

Determine the direction of motion.
The consecutive difference has the sign of \(2-x_n\), so the sequence is monotone.

Conclude and identify the limit.
Monotone boundedness gives convergence; the recurrence then forces the limit to be 2.

A common pitfall: a fixed point is not a convergence proof

Finding a fixed point answers which values could be limits if the sequence converges. It does not by itself establish that the sequence converges. A recurrence can have a fixed point while its terms fail to approach it. Here, convergence follows from the separate argument: an invariant bound together with monotonicity.

There is also a specific error to avoid when solving the fixed-point equation. Squaring \(L=\sqrt{2+L}\) gives two algebraic roots, but not every root of the squared equation satisfies the original equation. For example, substituting \(L=-1\) into the original gives \(-1=\sqrt{1}=1\), which is false. The limit must also be nonnegative because it is the limit of positive terms bounded on the nonnegative side. Therefore only \(L=2\) is possible.

The direction of motion matters as well. It would be incorrect to claim that the sequence always increases: the example with \(x_1=3\) decreases. The comparison with 2 determines the correct monotonicity, while the invariant interval ensures that the sequence does not cross to the other side.

Check Your Understanding

Use the invariant-interval and monotonicity argument to answer these questions.

For \(0<x_1<2\), which interval contains every term, and why does the sequence increase?
For \(x_1>2\), which interval contains every term, and why does the sequence decrease?
Show by factoring that the sign of \(x_{n+1}-x_n\) is determined by \(2-x_n\).
Which theorem establishes convergence after boundedness and monotonicity have been proved?
Why must the limit equation be checked against the original square-root equation after solving the squared equation?